text:
## Great Courses Launchpoint

The Teaching Company, LLC offers hundreds of video courses under the name “The Great Courses” on just about every subject imaginable, with more being added all the time.

Though offered for personal in-home viewing, these 30-minute lectures (or 45-minute in the case of Robert Greenberg’s engaging music courses) would make a wonderful centerpiece for a continuing education course.

As an instructor, what I would like to be able to do is show my class a Great Courses lecture, and then follow that with discussion and activities that reinforce and expand upon those  concepts during the remainder of a 60-minute or 90-minute class.

Not unlike what a good teaching assistant does in a college recitation section after a lecture by the professor, The Great Courses lecture would provide instructional scaffolding for both instructor and student.

I believe The Teaching Company has a great opportunity here.  Just by allowing an instructor to show a course to students (and charging a reasonable fee to do so), they would be opening up a new market for their products, and would no doubt bring in many new individual customers.

The Teaching Company could provide the courses “as is”, or could make available supplemental materials for the continuing education teacher and their students.

I even have a name for this new offering: Great Courses Launchpoint.

Currently, The Teaching Company doesn’t exactly encourage the use of their materials for face-to-face teaching:

My hope is that they will see the value of incorporating their video lectures into the classroom, and maybe Great Courses Launchpoint will roll out by the time I semi-retire in three or four years.  One of the frustrations of getting older is that my “day job” is taking a greater share of my time and available energy than ever before.  I love teaching, though, and semi-retirement will afford me the opportunity to begin teaching on a regular basis again.  Looking forward to it!

## Satellite, Meteor, and Aircraft Crossings 2018

Edmund Weiss (1837-1917) and many astronomers since have called asteroids “vermin of the sky”, but since October 4, 1957 another “species” of sky vermin made their debut: artificial satellites.  In the process of video recording stars for possible asteroid occultations, I frequently see satellites passing through my ~¼° field of view.

I’ve put together a video montage and some individual videos of satellites I’ve recorded between March 10, 2018 and November 24, 2018.  All of the events are shown below, with the boldface events being presented chronologically in the first video.  Both the NORAD and International designations are given for each satellite.  The range is the distance between observer and satellite at the time of observation.

UT Date
3-10-2018
3-25-2018
4-1-2018
4-2-2018
5-5-2018
7-6-2018
7-26-2018
7-31-2018
8-3-2018
8-23-2018
9-16-2018
10-21-2018 (2 satellites)
10-24-2018

Target Star
UCAC4 459-002239
TYC 621-45742-1
UCAC4 497-035454
UCAC4 416-092784
UCAC4 385-061427
N Sct 2018
UCAC4 429-110724
UCAC4 384-149264
UCAC4 362-194694
UCAC4 526-007192
UCAC4 316-210974
UCAC4 418-144100
UCAC4 302-215969

Satellite
SL-8 RB (Kosmos 726)
unknown space debris
unknown satellite
unknown satellite
unknown satellite
SL-8 RB (Kosmos 726)
Ariane 5 RB (VA209)
YURI 2A (BS-2A)
Kosmos 1092
SL-8 RB (Kosmos 80)
Galaxy 17 & NIMIQ 6
Sentinel 1B

Satellite
SL-8 RB (Kosmos 726)
unknown space debris
unknown satellite
unknown satellite
unknown satellite
SL-8 RB (Kosmos 726)
Ariane 5 RB (VA209)
YURI 2A (BS-2A)
Kosmos 1092
SL-8 RB (Kosmos 80)
Galaxy 17
NIMIQ 6
Sentinel 1B

Designation
7737; 1975-028-B
unknown
unknown
unknown
unknown
27946; 2003-043-B
7737; 1975-028-B
38780; 2012-051-C
14659; 1984-005-A
11326; 1979-030-A
1575; 1965-070-F
31307; 2007-016-B
38342; 2012-026-A
41456; 2016-025-A

Range & Direction
2,199.9 km SE
unknown SE
unknown SE
unknown NE
unknown NE
34,141.7 km NE
1,483.2 km SE
18,153.7 km NE
39,042.5 km NE
1,870.9 km NE
3,137.8 km NE
37,737.7 km E
37,736.3 km E
2,028.6 km NW

You’ll notice that sometimes the satellite crosses the field as a moving “dash”. That’s because sometimes I used longer exposure times to record a fainter target star.  A wind gust hit the telescope during the second event (3-25-2018).  The field is oriented North up and East to the left.  In this first video, you’ll notice that Sentinel 1B (the last event) has a unusual retrograde orbit (sun-synchronous) and is moving towards the NW.

In general, the slower the satellite is moving across the field, the higher is its orbit around the Earth.  One must also consider how much of the satellite’s orbital motion is along your line of sight to the satellite.

In the following video clip, you’ll see an unidentified piece of space debris, a very faint “dash” (due to integration) moving NE across the field from lower right to upper left, recorded on May 5, 2018 UT.

Next, we see a Ariane rocket body used to hoist SMART-1 towards the Moon and the Insat 3E and eBird 1 towards their geostationary orbits.  This recording was made on July 6, 2018 UT.  The rocket body is traveling NE (mostly east).  The light curve below the video suggests the possibility of some tumbling motion, but the satellite is faint and the photometry noisy.

And here is a rapidly tumbling (but low amplitude) Ariane rocket body, observed on July 31, 2018 UT and traveling NE.

Here is a no-longer-operational Japanese communications satellite named Yuri 2A, launched in 1984 and captured here on August 3, 2018 UT.  It is traveling NE (mostly east) and shows a beautiful long-period large-amplitude light curve.

Finally, we see not one but two geostationary communication satellites, Galaxy 17 (first and fainter) and NIMIQ 6 moving east across the field (as my telescope tracks westward to follow the Earth’s rotation), captured here on October 21, 2018 UT.  Galaxy 17 exhibits no discernible rotation, but NIMIQ 6 shows a low-amplitude long-period change in brightness.

Next we turn to three telescopic meteors I recorded on June 4, July 7, and September 11, 2018 UT.

UT Date
6-4-2018
7-7-2018
9-11-2018

Target Star
UCAC4 408-094611
UCAC4 275-188730
UCAC4 399-093188

Constellation & Direction
Scutum, SSE
Sagittarius, SW
Scutum, NNE

Here these meteors are presented in a video montage.

I even captured an airplane crossing the field on August 22, 2018 UT:

References
Hughes, D. W. & Marsden, B. G. 2007, J. Astron. Hist. Heritage, 10, 21

If you thought light pollution is bad (and it is!), radio pollution for radio astronomers is much worse.  Even years ago, terrestrial pollution of the radio spectrum tended to swamp faint celestial sources at many frequencies, and in 1958 the FCC established a 13,000 square mile rectangular region of West Virginia, Virginia, and Maryland as the National Radio Quiet Zone.  Two facilities within this protected region—whose natural topography helps to screen out many terrestrial radio emissions—are the Sugar Grove Station and the Green Bank Observatory near Green Bank, West Virginia.  The world’s largest fully-steerable radio telescope dish was built at Green Bank in 1956.  Though the original 300-ft. dish collapsed in 1988 due to a structural failure, it was rebuilt in 2000 as the Robert C. Byrd Green Bank Telescope, a leading facility for radio astronomy.

Counties wholly within the NRQZ, where many radio-emitting sources are regulated or banned outright, are Alleghany, Augusta, Bath, Highland, Nelson, and Rockbridge in Virginia, and Hardy, Pendleton, Pocahontas, Randolph, and Upshur in West Virginia.

The NRQZ isn’t the only radio quiet zone.  Here are some others:

• Arecibo Observatory, Puerto Rico
• Astronomy Geographic Advantage Act (AGAA), South Africa
• Atacama Large Millimeter Array (ALMA), Chile
• Five hundred meter Aperture Spherical Telescope (FAST), China
• Institute for Radio Astronomy in the Millimeter Range (IRAM), Spain
• Itapetinga Radio Observatory (IRO), Brazil
• Large Millimeter Telescope (LMT), Mexico
• Pushchino Radio Astronomy Observatory, Russia

The best place in the world to do radio astronomy is not on our world at all but instead on the far side of the Moon.  Radio telescopes deployed on the lunar farside could “listen” to the universe with absolutely no interference from Earth.  The solid body of the Moon (and its lack of an atmosphere) would completely block all radio signals and noise emanating from the Earth and Earth orbit.  And some radio telescopes could be quickly and easily deployed (think long-wire antennas rather than radio dishes).  Of course, the Moon itself will need to be designated as a radio quiet zone so that any lunar colonies, rovers, or satellites operate at frequencies and times that will not interfere with scientific work.  Maybe infrared or optical lasers would be a better way to communicate?

How would data from a lunar farside radio observatory be transmitted back to Earth?  One way would be to have a dedicated lunar satellite that receives data from the radio observatory while it is traveling over the lunar farside.  It would then re-transmit that data to Earth while it is traveling over the Earth-facing nearside.

Another (probably more expensive) approach would be to have a series of radio relay towers spaced at intervals from the radio observatory around to the lunar nearside where a transmitter could send the data back to Earth.

A third choice would be to locate the radio observatory in a libration zone along the border between the lunar nearside and farside.  At a libration zone radio observatory, data would be collected and stored until each time libration allows a direct line-of-sight to Earth.

The crater Daedalus, near the center of the lunar farside, has been suggested as the best location for a radio astronomy facility on the Moon (Pagana et al. 2006).

There is also a region above the farside lunar surface where radio emissions from Earth and Earth-orbiting satellites, would be blocked by the Moon, called the “Quiet Cone”, as illustrated in the diagram below.

The Earth-Moon L2 Lagrange point (EML2) is probably going to be within the lunar quiet cone.  Because L2 is an unstable Lagrange point, a radio telescope in the quiet cone would need to be in a halo orbit about EML2, and a tight one at that to avoid “seeing” any radio emissions from the highest Earth-orbiting satellites.

References
Antonietti, N.; Pagana, G.; Pluchino, S.; Maccone, C.
A proposed space mission around the Moon to measure the Moon Radio-Quiet Zone, 36th COSPAR Scientific Assembly. Held 16 – 23 July 2006, in Beijing, China.

## Name That Comet

As of this writing, there are 3,635 comets named SOHO, over 300 comets named LINEAR, some 179 comets named PANSTARRS, 82 comets named McNaught, 62 comets named NEAT, and so on.

Except for the comets discovered by Scottish-Australian astronomer Robert H. McNaught (1956-), all of the above comets were discovered by various automated surveys.

SOHO = Solar and Heliospheric Observatory (spacecraft)

LINEAR = Lincoln Near-Earth Asteroid Research

Pan-STARRS = Panoramic Survey Telescope and Rapid Response System

NEAT = Near-Earth Asteroid Tracking

How do we distinguish between comets having the same name?  Each has a separate comet designation.  The first Comet LINEAR has a designation of P/1997 A2, and the most recent Comet LINEAR has a designation of C/2017 B3.

A comet designation starts with one of the following prefixes:

P/ – a periodic comet (orbital period < 200 years or confirmed observations at more than one perihelion passage)

C/ – non-periodic comet (orbital period ≥ 200 years and confirmed observations at only one perihelion passage)

X/ – comet for which no reliable orbit could be calculated (generally, historical comets)

D/ – a periodic comet that has disappeared, broken up, or been lost

A/ – an object that was mistakenly identified as a comet, but is actually a minor planet (asteroid, trans-Neptunian object, etc.)

I/ – an interstellar object that did not originate in our solar system

This is then followed by the year of discovery, a letter indicating the half-month of discovery, followed by the numeric order of discovery during the half-month.

So, we can see that the first Comet LINEAR, P/1997 A2, is a periodic comet discovered in 1997, between January 1 and January 15 of that year, and it was the second comet to be discovered during that period of time.  After the second perihelion passage, P/1997 A2 (LINEAR) was subsequently given the periodic comet number prefix of 230, so the full designation for this comet is now 230P/1997 A2 (LINEAR).

Likewise, the most recent Comet LINEAR (at the time of this writing), C/2017 B3, is a non-periodic comet discovered in 2017 between January 16 and January 31, the third comet discovered during that period of time.

Interestingly, if different periodic comets have the same name, they are sequentially numbered.  Perhaps the most famous example is Comet Shoemaker-Levy 9 that broke up and crashed into Jupiter during July 1994.  There are a total of nine periodic comets named Shoemaker-Levy.  They are:

192P/1990 V1   Shoemaker-Levy 1
137P/1990 UL3  Shoemaker-Levy 2
129P/1991 C1   Shoemaker-Levy 3
118P/1991 C2   Shoemaker-Levy 4
145P/1991 T1   Shoemaker-Levy 5
181P/1991 V1   Shoemaker-Levy 6
138P/1991 V2   Shoemaker-Levy 7
135P/1992 G2   Shoemaker-Levy 8
D/1993 F2      Shoemaker-Levy 9

However, four additional non-periodic comets were discovered by the Carolyn & Gene Shoemaker and David Levy team.  They have not received a numeric suffix and are all called “Comet Shoemaker-Levy”:

C/1991 B1      Shoemaker-Levy
C/1991 T2      Shoemaker-Levy
C/1993 K1      Shoemaker-Levy
C/1994 E2      Shoemaker-Levy

This strikes me as a bit strange.  Why afford a numeric suffix to a comet name only when it is a periodic comet?  Why not give all comets named “Shoemaker-Levy” a numeric suffix.  Normally, we would number them all in order of discovery, but since the nine periodic comets have already received a number, we would have to number the four non-periodic comets as C/1991 B1 (Shoemaker-Levy 10), C/1991 T2 (Shoemaker-Levy 11), C/1993 K1 (Shoemaker-Levy 12), and C/1994 E2 (Shoemaker-Levy 13).

I would like to see all comets, both periodic and non-periodic, receive a numeric suffix to their names whenever there is more than one.  So, instead of Comet LINEAR we would have Comet LINEAR 1, Comet LINEAR 2, Comet LINEAR 3, and so on.

By the way, the days of amateur astronomers discovering a new comet will probably soon come to a close.  Though this is a little sad, it does tell us that the entire sky is being monitored much more closely than in the past, by a number of automated surveys.  And that is a good thing, because we will be much less likely to miss anything “new” in the sky.

None One of the comets this year (so far) have has been discovered by amateurs.

UPDATE – November 20, 2018: California amateur astronomer and prolific comet hunter Don Machholz, along with Japanese amateur astronomers Shigehisa Fujikawa and Masayuki Iwamoto, independently discovered a new comet on November 7.  The new long-period comet has been named C/2018 V1 (Machholz-Fujikawa-Iwamoto).  Remarkable!

Here is the current tally of comet discoveries (or recoveries) this year:

Pan-STARRS (Panoramic Survey Telescope and Rapid Response System)
C/2018 A1 (PANSTARRS)
364P/2018 A2 (PANSTARRS)
C/2018 A4 (PANSTARRS)
P/2018 A5 (PANSTARRS)
C/2018 F4 (PANSTARRS)
P/2018 H2 (PANSTARRS)
P/2018 L1 (PANSTARRS)
P/2018 L4 (PANSTARRS)
P/2018 P3 (PANSTARRS)
P/2018 P4 (PANSTARRS)
C/2018 P5 (PANSTARRS)
372P/2018 P6 (McNaught) [recovery of P/2008 O2]

ATLAS (Asteroid Terrestrial-impact Last Alert System)
C/2018 A3 (ATLAS)
C/2018 E1 (ATLAS)
C/2018 K1 (Weiland) [H. Weiland, ATLAS]
C/2018 L2 (ATLAS)
C/2018 O1 (ATLAS)

MLS (Mt. Lemmon Survey)
C/2018 A6 (Gibbs) [A.R. Gibbs, MLS]
C/2018 B1 (Lemmon)
C/2018 C2 (Lemmon)
C/2018 EF9 (Lemmon)  [originally classified as an asteroid]
C/2018 F1 (Grauer) [A.D. Grauer, MLS]
C/2018 F3 (Johnson) [J.A. Johnson, MLS]
C/2018 KJ3 (Lemmon) [originally classified as an asteroid]
P/2018 L5 (Leonard) [G. Leonard, MLS]
C/2018 R3 (Lemmon)
C/2018 R5 (Lemmon)

SONEAR (Southern Observatory for Near Earth Asteroid Research)
C/2018 E2 (Barros) [Joao Barros, SONEAR]

NEOWISE (Near-Earth Object Wide-field Infrared Survey Explorer)
C/2018 EN4 (NEOWISE)  [originally classified as a Centaur asteroid]
C/2018 N1 (NEOWISE)

Spacewatch
366P/2018 F2 (Spacewatch)

CSS (Catalina Sky Survey)
367P/2018 H1 (Catalina)
C/2018 M1 (Catalina)
C/2018 R4 (Fuls) [D.C. Fuls, CSS]

NEAT (Near-Earth Asteroid Tracking)
368P/2018 L3 (NEAT)
370P/2018 P2 (NEAT)

ASAS-SN (All Sky Automated Survey for SuperNovae)
C/2018 N2 (ASASSN)

OGS (ESA Optical Ground Station)
369P/2018 P1 (Hill) [recovery of P/2010 A1]
371P/2018 R1 (LINEAR-Skiff) [recovery of P/2001 R6]

373P/2018 R2 (Rinner)  [Jean-Francois Soulier, Maisoncelles, and Krisztian Sarneczky, University of Szeged, Piszkesteto Station (Konkoly), independently recovered P/2011 W2]

374P/2018 S1 (Larson) [Krisztian Sarneczky and Robert Szakats, University of Szeged, Piszkesteto Station (Konkoly), recovered P/2007 V1]

375P/2018 T1 (Hill) [Krisztian Sarneczky, University of Szeged, Piszkesteto Station (Konkoly), recovered P/2006 D1]

## Cosmology: The History and Nature of Our Universe

Mark Whittle, Professor of Astronomy at the University of Virginia, has put together the most comprehensive and comprehensible treatment on the subject of cosmology that I have ever encountered.  Cosmology: The History and Nature of Our Universe, a series of 36 thirty-minute video lectures for The Great Courses (Course No. 1830), is a truly remarkable achievement.

Even though this course was released ten years ago in 2008, all of the material is still completely relevant.  This is the course on cosmology that I’ve always wanted but never had.  Enjoy!

Cosmology has come a long ways since I was a physics and astronomy student at Iowa State University from 1975-1980, and again in 1981, 1984, and 2000-2005.  I’m glad to see a course specifically about cosmology is now offered at a number of universities.  When I was an undergraduate student at ISU, it was unheard of.  The University of Wisconsin at Madison Department of Astronomy currently offers both an undergraduate and a graduate course in cosmology: Astronomy 335 – Cosmology, and Astronomy 735 – Observational Cosmology.  And the Department of Physics & Astronomy at Iowa State University now offers an undergraduate/graduate dual-listed cosmology course: Astro 405/505 – Astrophysical Cosmology.

When I retire in a few years, I would love to be a “fly on the wall” at the UW-Madison astronomy department.  Wonder if they could use an expert SAS programmer to help analyze the massive quantities of data they surely must have?  (Though the last time I interviewed for an astronomy job, at the McDonald Observatory in Texas, the interviewers had never heard of SAS but asked if I knew Python, which of course is what nearly everyone is looking for and using these days.  Tomorrow, it will be something else…).  In retirement, at the very least I would love to immerse myself in a few astronomy courses at UW-Madison.  Something to look forward to!

## Blue Light Blues

One by one, all of our warm white lights are being replaced by cold, harsh, bluish-white LEDs.  And it is happening fast.

Everywhere.  In our streetlights, our workplaces, even our homes.  How do you like looking into those blue-white vehicle headlights as compared with the yellow-white ones we have been using since the automobile was invented?

LED lighting is the way of the future, don’t get me wrong, but we should be specifying and installing LED lights with a correlated color temperature (CCT) of 2700K or 3000K—with few exceptions—not the 4000K or higher that is the current standard.

Why is 4000K the current standard?  Because blue-white LEDs have a slightly greater luminous efficacy than yellow-white LEDs.  Luminous efficacy is the amount of light you get out for the power you put in, often measured in lumens per watt.  But should luminous efficiency be the only consideration?  What about aesthetics?  In addition to luminous efficacy, there are other, more significant ways to reduce power consumption and greenhouse gas emissions:

• Use the minimum amount of light needed for the application; no need to overlight
• Use efficient light fixtures that direct light only to where it is needed; near-horizontal light creates annoying and visibility-impairing glare and light trespass, and direct uplight into the night sky is a complete waste
• Produce the light only when it is needed through simple switches, time controls, and occupancy sensors; or, use lower light levels during times of little or no activity

Even the super-inefficient incandescent light bulb (with a CCT of 2400K, by the way), operating three hours each night uses less energy than the light source with the highest luminous efficacy operating dusk to dawn.  Think about it.

In my town, as in most now, the soothing orange 1900K high pressure sodium (HPS) streetlights are being replaced with 4000K LEDs.  That’s a big change.  It will completely transform our outdoor nighttime environment.  Warm-white compact fluorescents are 2700K, and even tungsten halogen bulbs are 3000K.  Do we really want or need 4000K+ LEDs?

We are currently witnessing a complete transformation of our illuminated built environment.  Not enough questions are being asked nor direction being given by citizens, employees, and municipalities.  The lighting industry generally wants to sell as many lights as possible at the highest profit margin.  We as lighting consumers need to make sure we have the right kind of light, the right amount of light, and lighting only when and where it is needed.

## Effective Diameter of an Irregularly-Shaped Object

A diameter of a circle in 2D is defined as any straight line segment that intersects the center of the circle with endpoints that lie on the circle.  Since all diameters of a circle have the same length, the diameter is the length of any diameter.

Likewise, a diameter of a sphere in 3D is defined as any straight line segment that intersects the center of the sphere with endpoints that lie on the surface of the sphere, and the diameter is its associated length.

But how do we define the diameter of an irregularly-shaped object such as a typical asteroid or trans-Neptunian object?

For a well-characterized object such as 951 Gaspra—the first asteroid to be photographed up close by a spacecraft—we’ll see the dimensions of the best fitting triaxial ellipsoid given in terms of “principal diameters”.  In the case of Gaspra, that is 18.2 × 10.5 × 8.9 km.

In certain circumstances, however, it would advantageous to characterize an irregularly-shaped object using a single “mean diameter”.  How should we calculate that?

There are two good approaches, provided you have enough information about the object.  The first is to determine the “volume equivalent diameter” which is the diameter of a sphere having the same volume as the asteroid.  This is particularly relevant to mass and density.

For purposes of illustration only, let’s assume Gaspra’s dimensions are exactly the same as its best-fitting triaxial ellipsoid.  If that were true, the volume of Gaspra would be

$V = \frac{{4\pi abc }}{3}$

where V is the volume, and a, b, and c are the principal radii of the triaxial ellipsoid.

Plugging in the numbers 9.1 km, 5.25 km, and 4.45 km (half the principal diameters), we get a volume of 890.5 km3.

The volume equivalent diameter is

$d_{vol} = \left (\frac{6V_{obj}}{\pi } \right )^{1/3}$

where dvol is the volume equivalent diameter, and Vobj is the volume of the object.

Plugging in the volume of 890.5 km3 gives us a volume equivalent diameter of 11.9 km.

The second approach is to determine the “surface equivalent diameter” which is the diameter of a sphere having the same surface area as the asteroid.  This is most relevant to reflectivity or brightness.

Once again using our triaxial ellipsoid as a stand-in for the real 951 Gaspra, we find that the general solution for the surface area of an ellipsoid requires the use of elliptic integrals.  However, there is an approximation that is more straightforward to calculate and accurate to within about 1%:

$S\approx 4\pi\left ( \frac{a^{p}b^{p}+a^{p}c^{p}+b^{p}c^{p}}{3} \right )^{1/p}$

where S is the surface area, p ≈ 1.6075 can be used, and a, b, and c are the principal radii of the triaxial ellipsoid.

Once again plugging in the numbers, we get a surface area of of 478.5 km2.

The surface equivalent diameter is

$d_{sur} = \left (\frac{S_{obj}}{\pi } \right )^{1/2}$

where dsur is the surface equivalent diameter, and Sobj is the surface area of the object.

Plugging in the surface area of 478.5 km3 gives us a surface equivalent diameter of 12.3 km.

You’ll notice that the surface equivalent diameter for 951 Gaspra (triaxial ellipsoid approximation) is 12.3 km which is larger than the volume equivalent diameter of 11.9 km.  The surface equivalent diameter is apparently always larger than the volume equivalent diameter, though I leave it as an exercise for the mathematically-inclined reader to prove that this is so.

References
Herald, David (2018, October 23).  [Online forum comment].  Message
posted to https://groups.yahoo.com/neo/groups/IOTAoccultations/conversations/messages/65158

Thomas, P.C., Veverka, J., Simonelli, D., et al.: 1994, Icarus 107The Shape of Gaspra, 23-26.

## Bike Path to Nowhere

The Dodgeville area is badly in need of an off-road paved (asphalt) bike path.  Every time I go to Madison, I am envious of all the bike trails they have.  Why can’t small towns like Dodgeville and rural areas have some paved bike paths, too?  Brigham County Park in rural Dane County has a beautiful new trail.  Why not Iowa County?

I’d really like to see the Military Ridge Trail between Dodgeville and Ridgeway paved.  Anyone interested in serving on an ad hoc committee with me to make that happen?

There is a 5.1-mile paved trail called the Shake Rag Trail which runs along US Highway 151 between Dodgeville and Mineral Point, but it is far from ideal.  First of all, there is no safe way to bike to it from Dodgeville!  You can ride through the hospital parking lot to Heritage Lane, head south until you get to Brennan Rd., turn right, but when you get to WI Highway 23, you have to ride along the east shoulder of that busy road with fast-moving vehicles for 0.4 miles to get to the bike path, as shown in the map below.

What a relief!  You’ve now reached the paved bike path, and it is off-road!

But, after traveling only 0.5 mile, the bike path suddenly ends at Chris-Na-Mar Road.

You now ride 0.7 miles on Chris-Na-Mar Road, and then the off-road bike path starts up again.

Now, you get to ride 1.3 miles on an off-road paved bike path.  Yay!  But the bike path again abruptly ends at County Road YD.  It is not clear what you should do next except maybe turn around?

Persistence pays off, and if you soldier on you’ll find that you can ride 2.1 miles on County Road YD until you reach the off-road bike path again.  You’re almost to Mineral Point!

The bike path goes another 0.5 mile until it ends at Shakerag St. in Mineral Point.  You’ve traveled a total of 5.1 miles on the Shake Rag Trail, but less than half of it was on a bona fide bike path.

Don’t get me wrong, I’m really glad that the Shake Rag Trail got built.  But for any of you who have ridden the crushed rock Military Ridge Trail between Dodgeville and Ridgeway (all off-road), you’ll understand how much nicer Military Ridge Trail would be than the Shake Rag Trail if only it were paved.

## Eclipsing Binaries

With the advent of relatively inexpensive CCD cameras, amateur astronomers with modest-sized telescopes are in an excellent position to contribute valuable scientific data to the astronomical community.  One type of object that can be very interesting and useful to observe is the eclipsing binary.  And there are a lot of them.

Due to a sometimes fortuitous alignment of the orbital plane of a binary star along or near our line of sight, one or both stars pass directly in front of the other periodically, and this type of object is known as an eclipsing binary.

The brightest eclipsing binary in our sky is Algol (Beta (β) Persei).  Known to vary in brightness since antiquity, astute ancient Arab astronomers gave Beta Persei the name “al Ghul” which, loosely translated, means “the Demon Star”.  Today, we know that Algol’s brightness variations are caused by a hot blue B8V star (Algol A) going behind and in front of its cooler and less massive but larger K0IV companion (Algol B).  Since the two stars orbit each other once every 2.867328 days (they are very close, separated by just a little over 5½ million miles), every 2 days, 20 hours, 48 minutes, and 57 seconds Algol B passes in front of much-brighter Algol A for a few hours, and the single point of light we see from Earth dims by 1.3 magnitudes.  This is the primary eclipse.  A secondary eclipse also occurs half a period before or after each primary eclipse.  When Algol A passes in front of Algol B, the brightness of the point of light we see drops by only 0.05 magnitude.  This shallow secondary minimum occurs because Algol B is not nearly as bright as Algol A.

Eclipsing binaries like Algol (which are close enough to each other to form an interacting pair) are interesting subjects for amateur astronomers to monitor.  Periods can change, phases can shift, and unexpected events can occur, such as when Dr. Jim Pierce (now Emeritus Professor of Astronomy at Minnesota State University in Mankato) and I were the first to observe ultraviolet flare events from the eclipsing binary V471 Tau at Iowa State University’s Erwin W. Fick Observatory in 1978.

So, how do you know when eclipses will occur, how deep they will be, and how long to monitor the star before, during, and after the event?  A great starting point is the Eclipsing Binary Ephemeris Generator by Shawn Dvorak which shows you a number of stars that will be in eclipse and observable from your location on any given night.  The Timing Database at Krakow (TIDAK), maintained by Jerzy M. Kreiner at the Mt. Suhora Astronomical Observatory in Poland, is another great source of eclipsing binary information.

A schedule, if you will, of eclipsing binary primary eclipses (like other astronomical events) is called an ephemeris.  Eclipsing binary ephemerides look like this one for Algol:

HJD = 2452500.21 + E × 2.867315

Here, HJD is the heliocentric Julian date of minimum light.  Julian date is a continuous count of days and fractions thereof elapsed since an arbitrary starting date of noon Universal Time (UT) on January 1, 4713 B.C.  The heliocentric Julian date removes the orbital motion of the Earth from the ephemeris calculations, centering the times of events on the Sun rather than the Earth.  An event could be observed to occur as much as 8.3 minutes earlier or later than calculated depending on where the Earth is in her orbit relative to the star.  The first number in the equation above, in this case 2452500.21, refers to the heliocentric Julian date of some arbitrary starting minimum.  The E stands for epoch, simply a consecutive integer count of successive minima, and the second number, in this case 2.867315, refers to the orbital period of the eclipsing binary in days.  The Kreiner website takes the chore out of choosing the appropriate value of E for the time you want to observe by calculating the HJDs (and corresponding Earth-based UT dates and times) of the eclipsing binary you choose over the next several days.

You should monitor a star before, during, and after the eclipse, so having a rough of idea of what object you should observe and when does not require you convert heliocentric Julian date to the Julian date at the telescope. However, any event times from data you record at the telescope must be converted to HJD for it to be useful.  There is an online tool to do this for you.  Of course, you not only need to know the UT date and time of an event, but also the equatorial coordinates (right ascension and declination) of the object you were observing to calculate the heliocentric Julian date.

We’re not even going to get into barycentric Julian date (BJD), or the fact that the distance between the Sun (or the barycenter of the solar system) and the eclipsing binary of interest is growing (radial velocity > 0) or shrinking (radial velocity < 0), and that this means that the period we measure is not exactly the same as the true orbital period of the system.  But it is very close.


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# Centroid of a triangle definition and formula explained

Learn and know about the centroid of a triangle. Centroid is one of the most important topics one should study in coordinate geometry chapter. So now we will learn the actual meaning of the centroid and also we will learn how we can find the centroid.

First, we will see the definition of the centroid and after that, we will see properties of centroid, formula and representation of centroid for different kinds of triangle one by one.

## Definition of the centroid of a triangle:

The intersection point of all the three medians of a triangle is called centroid. Centroid is represented with the letter G.

In the above triangle, we can observe three medians i.e. AD, BE and CF. we can also observe that all the three medians are meeting at one point, that point we are going to call as the centroid ( G).

Important Property of a centroid:

We should know that centroid (G ) divides the medians in 2: 1 ratio.

### The Formula for finding the centroid of a triangle with vertices:

Suppose If A $\left( { x }_{ 1 },{ y }_{ 1 } \right)$ , B $\left( { x }_{ 2 },{ y }_{ 2 } \right)$ C $\left( { x }_{ 3 },{ y }_{ 3 } \right)$ are the given three vertices of a triangle then centroid is given by

#### Problems based on the centroid:

Find the centroid of triangle with vertices A (1, 5), B (2, 7) and C (6, 3).

Solution:

In a Δ ABC, the centroid is (7, 2) and the two vertices of the triangle are given by (1, 8), (10, 4) then find the third vertex of a triangle.

Solution:

Let the two vertices of a triangle be A (1, 8), B (10, 4) and the third vertex be C (x, y).

Representation of centroid for right, acute and obtuse angled triangles:

I hope you learnt everything about the centroid i.e. definition, formula and property.

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Question 1: Which of the following cannot be valid assignment of probability for elementary events or
outcomes of sample space $S = \{ w_1, w_2, w_3, w_4, w_5, w_6, w_7 \}$

 Elementary Events $w_1$$w_1$ $w_2$$w_2$ $w_3$$w_3$ $w_4$$w_4$ $w_5$$w_5$ $w_6$$w_6$ $w_7$$w_7$ (i) $0.1$$0.1$ $0.01$$0.01$ $0.05$$0.05$ $0.03$$0.03$ $0.01$$0.01$ $0.2$$0.2$ $0.6$$0.6$ (ii) $\displaystyle \frac{1}{7}$$\displaystyle \frac{1}{7}$ $\displaystyle \frac{1}{7}$$\displaystyle \frac{1}{7}$ $\displaystyle \frac{1}{7}$$\displaystyle \frac{1}{7}$ $\displaystyle \frac{1}{7}$$\displaystyle \frac{1}{7}$ $\displaystyle \frac{1}{7}$$\displaystyle \frac{1}{7}$ $\displaystyle \frac{1}{7}$$\displaystyle \frac{1}{7}$ $\displaystyle \frac{1}{7}$$\displaystyle \frac{1}{7}$ (iii) $0.7$$0.7$ $0.6$$0.6$ $0.5$$0.5$ $0.4$$0.4$ $0.3$$0.3$ $0.2$$0.2$ $0.1$$0.1$ (iv) $\displaystyle \frac{1}{14}$$\displaystyle \frac{1}{14}$ $\displaystyle \frac{2}{14}$$\displaystyle \frac{2}{14}$ $\displaystyle \frac{3}{14}$$\displaystyle \frac{3}{14}$ $\displaystyle \frac{4}{14}$$\displaystyle \frac{4}{14}$ $\displaystyle \frac{5}{14}$$\displaystyle \frac{5}{14}$ $\displaystyle \frac{6}{14}$$\displaystyle \frac{6}{14}$ $\displaystyle \frac{15}{14}$$\displaystyle \frac{15}{14}$

For each event to be a valid assignment of probability, the probability of each event in sample space should be less than 1 and the sum of probability of all the events should be exactly equal to 1

(i) it is valid as each $P(w_i) ( \ for \ i=1 \ to \ 7)$ lies between 0 to 1 and sum of $P(w_1) = 1$

(ii) it is valid as each $P(w_i) ( \ for \ i = 1 \ to \ 7)$ lies between 0 to 1 and sum of $P(w_1) =1$

(iii) it is not valid as sum of $P(w_i)=2.8$ which is greater than 1

(iv) it is not valid as $\displaystyle P(w_7) = \frac{15 }{14 }$ which is greater than 1.

$\\$

Question 2: A die is thrown. Find the probability of getting:
(i) a prime number           (ii) 2 or 4           (iii) a multiple of 2 or 3.

(iv) an even prime number          (v) a number greater than 5

(vi) a number lying between 2 and 6

(i) Prime numbers on a dice are 2, 3, and 5. So, the total number of prime numbers is 3. We know that, Probability = Number of favorable outcomes/ Total number of outcomes Thus, probability of getting a prime number $\displaystyle = \frac{3 }{6} = \frac{1}{2}$

(ii) For getting 2 and 4, clearly the number of favorable outcomes is 2. We know that Probability = Number of favorable outcomes/ Total number of  $\displaystyle \text{outcomes Thus, the probability of getting 2 or 4} = \frac{2 }{6} = \frac{1}{3}$

(iii) Multiple of 2 are 3 are 2, 3, 4 and 6. So, the number of favorable outcomes is 4 We know that, Probability = Number of favorable outcomes/ Total number  $\displaystyle \text{of outcomes Thus, the probability of getting an multiple of 2 or 3} = \frac{4 }{6} = \frac{2}{3}$

(iv) An even prime number is 2 only. So, the number of favorable outcomes is 1. We know that, Probability = Number of favorable outcomes/ Total number of $\displaystyle \text{outcomes Thus, the probability of getting an even prime number }= \frac{1}{6}$

(v) A number greater than 5 is 6 only. So, the number of favorable outcomes is 1. We know that, Probability = Number of favorable outcomes/ Total number of  $\displaystyle \text{outcomes Thus, the probability of getting a number greater than 5} = \frac{1}{6}$

(vi) Total number on a dice is 6. Numbers lying between 2 and 6 are 3, 4 and 5 So, the total number of numbers lying between 2 and 6 is 3. We know that, Probability = Number of favorable outcomes/ Total number of outcomes Thus, $\displaystyle \text{the probability of getting a number lying between 2 and 6 } = \frac{3}{6} = \frac{1}{2}$

$\\$

Question 3: In a simultaneous throw of a pair of dice, find the probability of getting:

(i) 8 as the sum     (ii) a doublet     (iii) a doublet of prime numbers

(iv) a doublet of odd numbers     (v) a sum greater than 9

(vi) an even number on first

(vii) an even number on one and a multiple of 3 on the other

(viii) neither 9 nor 11 as the sum of the numbers on the faces

(ix) a sum less than 6     (x) a sum less than 7     (xi) a sum more than 7

(xii) neither a doublet nor a total of 10

(xiii) odd number on the first and 6 on the second

(xiv) a number greater than 4 on each die     (xv) a total of 9 or 11

(xvi) a total greater than 8

In a throw of pair of dice, total no of possible outcomes $= 36 (6 \times 6)$ which are

$(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 1), (2, 2), (2, 3) ,(2, 4), (2, 5), (2, 6)$

$(3, 1), (3, 2) ,(3, 3), (3, 4) ,(3, 5), (3, 6), (4, 1), (4, 2), (4, 3) ,(4, 4), (4, 5), (4, 6)$

$(5, 1), (5, 2), (5, 3) ,(5, 4), (5, 5), (5, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)$

(i) Let E be event of getting the sum as 8

No. of favorable outcomes $= 5 \text{ i.e. } \{ (2, 6) , (3, 5) , (4, 4) , (5, 3) , (6, 2) \}$

$\displaystyle \text{Probability } P(E) = \frac{ (\text{No. of favorable outcomes})}{(\text{Total no. of possible outcomes}) } = \frac{5}{36}$

(ii) Let E be the event of getting a doublet

No. of favorable outcomes $= 5 \text{ i.e. } \{ (1, 1) , (2, 2) , (3, 3) , (4, 4) , (5, 5) , (6, 6) \}$

$\displaystyle \text{Probability } P(E) = \frac{ (\text{No. of favorable outcomes})}{(\text{Total no. of possible outcomes}) } = \frac{6}{36} = \frac{1}{6}$

(iii) Let E be the event of getting a doublet of prime no’s

No. of favorable outcomes $= 3 \text{ i.e. } \{ (2, 2) , (3, 3) , (5, 5) \}$

$\displaystyle \text{Probability } P(E) = \frac{ (\text{No. of favorable outcomes})}{(\text{Total no. of possible outcomes}) } = \frac{3}{36} = \frac{1}{12}$

(iv) Let E be the event of getting a doublet of odd no’s

No. of favorable outcomes $= 3 \text{ i.e. } \{ (1, 1) , (3, 3) , (5, 5) \}$

$\displaystyle \text{Probability } P(E) = \frac{ (\text{No. of favorable outcomes})}{(\text{Total no. of possible outcomes}) } = \frac{3}{36} = \frac{1}{12}$

(v) E ⟶ event of getting a sum greater than 9

No. of favorable outcomes $= 6 \text{ i.e. } \{ (4, 6) , (5, 5) , (5, 6) , (6, 4) , (6, 5) , (6, 6) \}$

$\displaystyle \text{Probability } P(E) = \frac{ (\text{No. of favorable outcomes})}{(\text{Total no. of possible outcomes}) } = \frac{6}{36} = \frac{1}{6}$

(vi) Let E be the event of getting an even no. on first

No. of favorable outcomes $= 18 \text{ i.e. } \{(2, 1) , (2, 2) , (2, 3) , (2, 4) , (2, 5) , (2, 6) , (4, 1) , (4, 2) , (4, 3) , \\ (4, 4) , (4, 5) , (4, 6) , (6, 1) , (6, 2) , (6, 3) , (6, 4) , (6, 5) , (6, 6) \}$

$\displaystyle \text{Probability } P(E) = \frac{ (\text{No. of favorable outcomes})}{(\text{Total no. of possible outcomes}) } = \frac{18}{36} = \frac{1}{3}$

(vii) Let E be the event of getting an even no. on one and a multiple of 3 on other

No. of favorable outcomes $= 11 \text{ i.e. } \{ (2, 3) , (2, 6) , (4, 3) , (4, 6) , (6, 3) , (6, 6) , (3, 2) , (3, 4) , (3, 4) , (3, 6) , (6, 2) , (6, 4) \}$

$\displaystyle \text{Probability } P(E) = \frac{ (\text{No. of favorable outcomes})}{(\text{Total no. of possible outcomes}) } = \frac{11}{36}$

(viii) Let $\overline{E}$ be the event of getting neither 9 nor 11 as the sum of numbers on faces

Therefore E is  getting either 9 or 11 as the sum of no’s on faces

No. of favorable outcomes $= 6 \text{ i.e. } \{ (3, 6) , (4, 5) , (5, 4) , (6, 3) , (5, 6) , (6, 5) \}$

$\displaystyle \text{Probability } P(E) = \frac{ (\text{No. of favorable outcomes})}{(\text{Total no. of possible outcomes}) } = \frac{6}{36} = \frac{1}{6}$

$\displaystyle P(\overline{E}) = 1 - P(E) = 1 - \frac{1}{6} = \frac{5}{6}$

(ix) Let E be the event of getting a sum less than 6

No. of favorable outcomes $= 10 \text{ i.e. } \{(1, 1) , (1, 2) , (1, 3) , (1, 4) , (2, 1) , (2, 2) , (2, 3) , (3, 1) , (3, 2) , (4, 1)\}$

$\displaystyle \text{Probability } P(E) = \frac{ (\text{No. of favorable outcomes})}{(\text{Total no. of possible outcomes}) } = \frac{10}{36} = \frac{5}{18}$

(x) Let E be the event of getting a sum less than 7

No. of favorable outcomes $= 15 \text{ i.e. } \{(1, 1) , (1, 2) , (1, 3) , (1, 4) , (1, 5) , (2, 1) , (2, 2) , (2, 3) , (2, 4) , \\ (3, 1) , (3, 2) , (3, 3) , (4, 1) , (4, 2) , (5, 1)\}$

$\displaystyle \text{Probability } P(E) = \frac{ (\text{No. of favorable outcomes})}{(\text{Total no. of possible outcomes}) } = \frac{15}{36} = \frac{5}{12}$

(xi) Let E be the event of getting a sum more than 7

No. of favorable outcomes $= 15 \text{ i.e. } \{(2, 6) , (3, 5) , (3, 6) , (4, 4) , (4, 5) , (4, 6) , (5, 3) , (5, 4) , (5, 5) , \\ (5, 6) , (6, 2) , (6, 3) , (6, 4) , (6, 5) , (6, 6)\}$

$\displaystyle \text{Probability } P(E) = \frac{ (\text{No. of favorable outcomes})}{(\text{Total no. of possible outcomes}) } = \frac{15}{36} = \frac{5}{12}$

(xii) Let E be the event of getting a 1 at least once

No. of favorable outcomes $= 11 \text{ i.e. } \{(1, 1) , (1, 2) , (1, 3) , (1, 4) , (1, 5) , (1, 6) , (2, 1) , (3, 1) , (4, 1) , (5, 1) , (6, 1)\}$

$\displaystyle \text{Probability } P(E) = \frac{ (\text{No. of favorable outcomes})}{(\text{Total no. of possible outcomes}) } = \frac{11}{36}$

(xiii) Let E be the event of getting a no other than 5 on any dice

No. of favorable outcomes $= 25 \text{ i.e. } \{(1, 1) , (1, 2) , (1, 3) , (1, 4) , (1, 6) , (2, 1) , (2, 2) , (2, 3) , (2, 4) , (2, 6) , (3, 1) , \\ (3, 2) , (3, 3) , (3, 4) , (3, 6) , (4, 1) , (4, 2) , (4, 3) , (4, 4) , (4, 6) , (6, 1) , (6, 2) , (6, 3) , (6, 4) , (6, 6)\}$

$\displaystyle \text{Probability } P(E) = \frac{ (\text{No. of favorable outcomes})}{(\text{Total no. of possible outcomes}) } = \frac{25}{36}$

$\\$

Question 4: In a single throw of three dice, find the probability of getting a total of 17 or 18.

Three dices are thrown once.

Total number of possible outcomes is $\displaystyle 6^3 = 216$

$\displaystyle \text{Therefore, } n (S) = 216$

Let E be the event of getting total of 17 or 18

$\displaystyle \text{Therefore } E = 4 \text{ i.e.} \{ (6, 6, 5), (6, 5, 6), (5, 6, 6), (6, 6, 6) \}$

$\displaystyle n (E) = 4$

$\displaystyle P (E) = \frac{n (E) }{ n (S)} = \frac{4 }{ 216} = \frac{1}{54}$

$\\$

Question 5: Three coins are tossed together. Find the probability of getting:
(i) exactly two heads        (ii) at least two heads       (iii) at least one head and one tail.

Three coins are tossed together

Total number of possible outcomes is $\displaystyle 2^3 = 8$

$\displaystyle \text{Therefore, } n (S) = 8$

Sample space for three coins tossed is  $\displaystyle \{ HHH, HHT, HTH, THH, HTT, THT, TTH, TTT \}$

$\displaystyle n (E) = 3$

$\displaystyle P (E) = \frac{n (E) }{ n (S)} = \frac{3 }{ 8}$

$\displaystyle n (E) = 4$

$\displaystyle P (E) = \frac{n (E) }{ n (S)} = \frac{4 }{ 8} = \frac{1}{2}$

(iii) at least one head and one tail.

$\displaystyle n (E) = 4$

$\displaystyle P (E) = \frac{n (E) }{ n (S)} = \frac{4 }{ 8} = \frac{1}{4}$

$\\$

Question 6: What is the probability that an ordinary year has 53 Sundays?

Ordinary year has 365 days

365 days = 52 weeks + 1 day

That 1 day may be Sunday, Monday, Tuesday, Wednesday, Thursday, Friday, Saturday

Total no. of possible outcomes $\displaystyle n(S) = 7$

Let E be the event of getting 53 Sundays

No. of favorable outcomes = 1 {Sunday}

$\displaystyle \text{No. of favorable outcomes} n (E) = 1$

$\displaystyle P (E) = \frac{n (E) }{ n (S)} = \frac{1 }{ 7}$

$\\$

Question 7: What is the probability that a leap year has 53 Sundays and 53 Mondays?

A leap year consists of 366 days comprising of 52 weeks and 2 days.

There are 7 possibilities for these 2 extra days viz.

(i) Sunday, Monday     (ii) Monday, Tuesday     (iii) Tuesday, Wednesday,

(iv) Wednesday, Thursday     (v) Thursday, Friday     (vi) Friday, Saturday and

(vii) Saturday, Sunday.

Let us consider two events :

A : the leap year contains 53 Sundays

B : the leap year contains 53 Mondays.

$\displaystyle \text{Then we have } P(A) = \frac{2}{7}, \ \ \ P(B) = \frac{2}{7}$

$\displaystyle P(A \cap B) = \frac{1}{7}$

Hence, required probability $\displaystyle = P(A \cup B) = P(A) + P(B) - P(A \cap B) = \frac{2}{7} + \frac{2}{7} - \frac{1}{7} = \frac{3}{7}$

$\\$

Question 8: A bag contains 8 red and 5 white balls. Three balls are drawn at random. Find the probability that: (i) All the three balls are white.      (ii) A11 the three balls are red.        (iii) One ball is red and two balls are white

A bag contains 8 red and 5 white balls.

$\displaystyle \text{ Total number of possible outcomes } n(S) = ^{13}C_{3}$

(i) Probability when all the three balls are white

$\displaystyle n(E) = ^{5}C_{3}$

$\displaystyle P (E) = \frac{n (E) }{ n (S)} = \frac{^{5}C_{3} }{ ^{13}C_{3}} = \frac{ \frac{5!}{3! 2!} }{ \frac{13!}{3! 10!} } = \frac{10}{ \frac{13 \times 12 \times 11}{6} } = \frac{5}{143}$

(ii) Probability when all the three balls are red

$\displaystyle n(E) = ^{8}C_{3}$

$\displaystyle P (E) = \frac{n (E) }{ n (S)} = \frac{^{8}C_{3} }{ ^{13}C_{3}} = \frac{ \frac{8!}{5! 3!} }{ \frac{13!}{3! 10!} } = \frac{ \frac{8 \times 7 \times 6}{6} }{ \frac{13 \times 12 \times 11}{6} } = \frac{28}{143}$

(iii) Probability when One ball is red and two balls are white

$\displaystyle n(E) = ^{8}C_{1} \times ^{5}C_{2}$

$\displaystyle P (E) = \frac{n (E) }{ n (S)} = \frac{^{8}C_{1} \times ^{5}C_{2} }{ ^{13}C_{3}} = \frac{ \frac{8!}{7! 1!} \times \frac{5!}{3! 2!} }{ \frac{13!}{3! 10!} } = \frac{ 8 \times 10 }{ \frac{13 \times 12 \times 11}{6} } = \frac{40}{143}$

$\\$

Question 9: In a single throw of three dice, find the probability of getting the same number on all the three dice.

Three dices are thrown once.

Total number of possible outcomes is $\displaystyle 6^3 = 216$

$\displaystyle \text{Therefore, } n (S) = 216$

Let E be the event of getting same number on all the three dice

$\displaystyle \text{Therefore } E = 6 \text{ i.e.} \{(1,1,1), (2,2,2), (3,3,3), (4,4,4), (5,5,5) (6,6,6) \}$

$\displaystyle n (E) =6$

$\displaystyle P (E) = \frac{n (E) }{ n (S)} = \frac{6 }{ 216} = \frac{1}{36}$

$\\$

Question 10: Two unbiased dice are thrown. Find the probability that the total of the numbers on the dice is greater than 10.

Two dices are thrown.

Total number of possible outcomes is $\displaystyle 6^2 = 36$

$\displaystyle \text{Therefore, } n (S) = 36$

Let E be the event of getting sum on dice greater than 10

$\displaystyle \text{Therefore } E = 3 \text{ i.e.} \{ (5, 6), (6, 5), (6, 6) \}$

$\displaystyle n (E) =3$

$\displaystyle P (E) = \frac{n (E) }{ n (S)} = \frac{3 }{ 36} = \frac{1}{12}$

$\\$

Question 11: A card is drawn at random from a pack of 52 cards. Find the probability that the card drawn
15:

(i) a black king      (ii) either a black card or a king      (iii) black and a king

(iv) a jack, queen or a king      (v) neither a heart nor a king

(vi) spade or an ace     (vii) neither an ace nor a king

(viii) neither a red card nor a queen       (ix) other than an ace

(x) a ten      (xi) a spade      (xii) a black card      (xiii) the seven of clubs

(xiv) jack      (xv) the ace of spades      (xvi) a queen      (xvii) a heart

(xviii) a red card

A card is drawn at random from a pack of 52 cards.

Total number of possible outcomes is $\displaystyle = 52$

$\displaystyle \text{Therefore, } n (S) = 52$

(i) a black king

Let E be the event of of getting a black king.

$\displaystyle \text{Therefore } E = 2 \text{ i.e.} \{ \text{King of Spade, King of Clubs} \}$

$\displaystyle n (E) =2$

$\displaystyle P (E) = \frac{n (E) }{ n (S)} = \frac{2 }{ 52} = \frac{1}{26}$

(ii) either a black card or a king

Let E be the event of of getting either a black card or a king

$\displaystyle \text{Therefore } E = 28 \text{ i.e.} \{ \text{13 spades, 13 clubs, king of hearts and diamonds} \}$

$\displaystyle n (E) =28$

$\displaystyle P (E) = \frac{n (E) }{ n (S)} = \frac{28 }{ 52} = \frac{7}{13}$

(iii) black and a king

Let E be the event of of getting either a black and a king

$\displaystyle \text{Therefore } E = 2 \text{ i.e.} \{ \text{king of spade and king of clubs} \}$

$\displaystyle n (E) =2$

$\displaystyle P (E) = \frac{n (E) }{ n (S)} = \frac{2 }{ 52} = \frac{1}{26}$

(iv) a jack, queen or a king

Let E be the event of of getting a jack, queen or a king

$\displaystyle \text{Therefore } E = 12 \text{ i.e.} \{ \text{4 jacks, 4 queens and 4 kings} \}$

$\displaystyle n (E) =12$

$\displaystyle P (E) = \frac{n (E) }{ n (S)} = \frac{12 }{ 52} = \frac{3}{13}$

(v) neither a heart nor a king

Let E be the event of of getting neither a heart nor a king

$\displaystyle \text{Therefore } E = 52 - 13 - 3 = 36 \{ \text{since we have 13 hearts, 3 kings each of spades, clubs and diamonds} \}$

$\displaystyle n (E) =36$

$\displaystyle P (E) = \frac{n (E) }{ n (S)} = \frac{36 }{ 52} = \frac{9}{13}$

Let E be the event of of getting spade or an ace

$\displaystyle \text{Therefore } E = 16 \{ \text{13 spades and 3 aces each of hearts, diamonds and clubs} \}$

$\displaystyle n (E) =16$

$\displaystyle P (E) = \frac{n (E) }{ n (S)} = \frac{16 }{ 52} = \frac{4}{13}$

(vii) neither an ace nor a king

Let E be the event of of getting neither an ace nor a king

$\displaystyle \text{Therefore } E = 52-4-4 = 44 \{ \text{Since we have 4 aces and 4 kings} \}$

$\displaystyle n (E) =44$

$\displaystyle P (E) = \frac{n (E) }{ n (S)} = \frac{44 }{ 52} = \frac{11}{13}$

(viii) neither a red card nor a queen

Let E be the event of of getting neither a red card nor a queen

$\displaystyle \text{Therefore } E = 52 - 26 - 2 = 24 \{ \text{ Since we have 26 red cards of hearts and } \\ \text{ diamonds and 2 queens each of heart and diamond } \}$

$\displaystyle n (E) =24$

$\displaystyle P (E) = \frac{n (E) }{ n (S)} = \frac{24 }{ 52} = \frac{6}{13}$

(ix) other than an ace

Let E be the event of of getting other than an ace

$\displaystyle \text{Therefore } E = 52-4 = 48 \{ \text{Since we have 4 ace cards} \}$

$\displaystyle n (E) =48$

$\displaystyle P (E) = \frac{n (E) }{ n (S)} = \frac{48 }{ 52} = \frac{12}{13}$

(x) a ten

Let E be the event of of getting a ten

$\displaystyle \text{Therefore } E = 4 \{ \text{10 of spades, clubs, diamonds and hearts } \}$

$\displaystyle n (E) =4$

$\displaystyle P (E) = \frac{n (E) }{ n (S)} = \frac{4 }{ 52} = \frac{1}{13}$

Let E be the event of of getting a spade

$\displaystyle \text{Therefore } E = 13 \{ \text{13 Spade cards} \}$

$\displaystyle n (E) =13$

$\displaystyle P (E) = \frac{n (E) }{ n (S)} = \frac{13 }{ 52} = \frac{1}{4}$

(xii) a black card

Let E be the event of of getting a black card

$\displaystyle \text{Therefore } E = 26 \{ \text{13 Spade cards and 13 Club cards} \}$

$\displaystyle n (E) =26$

$\displaystyle P (E) = \frac{n (E) }{ n (S)} = \frac{26 }{ 52} = \frac{1}{2}$

(xiii) the seven of clubs

Let E be the event of of getting the seven of clubs

$\displaystyle \text{Therefore } E = 1 \{ \text{7 of Clubs} \}$

$\displaystyle n (E) =1$

$\displaystyle P (E) = \frac{n (E) }{ n (S)} = \frac{1 }{ 52}$

(xiv) jack

Let E be the event of of getting jack

$\displaystyle \text{Therefore } E = 4 \{ \text{4 Jacks} \}$

$\displaystyle n (E) =4$

$\displaystyle P (E) = \frac{n (E) }{ n (S)} = \frac{4 }{ 52} = \frac{1}{13}$

Let E be the event of of getting the ace of spades

$\displaystyle \text{Therefore } E = 1 \{ \text{Ace of Spade} \}$

$\displaystyle n (E) =1$

$\displaystyle P (E) = \frac{n (E) }{ n (S)} = \frac{1 }{ 52}$

(xvi) a queen

Let E be the event of of getting a queen

$\displaystyle \text{Therefore } E = 4 \{ \text{4 Queens} \}$

$\displaystyle n (E) =4$

$\displaystyle P (E) = \frac{n (E) }{ n (S)} = \frac{4 }{ 52} = \frac{1}{13}$

(xvii) a heart

Let E be the event of of getting a heart

$\displaystyle \text{Therefore } E = 13 \{ \text{13 hearts } \}$

$\displaystyle n (E) =13$

$\displaystyle P (E) = \frac{n (E) }{ n (S)} = \frac{13 }{ 52} = \frac{1}{4}$

(xviii) a red card

Let E be the event of of getting a red card

$\displaystyle \text{Therefore } E = 26 \{ \text{13 Diamonds and 13 Hearts} \}$

$\displaystyle n (E) =26$

$\displaystyle P (E) = \frac{n (E) }{ n (S)} = \frac{26 }{ 52} = \frac{1}{2}$

$\\$

Question 12: In shuffling a pack of 52 playing cards, four are accidently dropped; find the chance that the missing cards should be one from each suit.

We know that, from well shuffled pack of cards, 4 cards missed out total possible outcomes are

$\displaystyle \text{Therefore, } n (S) = ^{52} C_{4} = 270725$

Let E be the event that four missing cards are from each suite

$\displaystyle n(E) = ^{13} C_{1} \times ^{13} C_{1} \times ^{13} C_{1} \times ^{13} C_{1} = 13^4$

$\displaystyle P (E) = \frac{n (E) }{ n (S)} = \frac{13^4 }{270725} = \frac{2197}{20825}$

$\\$

Question 13: From deck of 52 cards, four cards are drawn simultaneously, find the chance that they will be the four honors of the same suit.

$\displaystyle n (S) = ^{52} C_{4} = 270725$

Let E be the event that all the cards drawn are face cards of same suit.

$\displaystyle n(E) = 4 \times ^{4} C_{4} = 4$

$\displaystyle P (E) = \frac{n (E) }{ n (S)} = \frac{4 }{270725}$

$\\$

Question 14: Tickets numbered from 1 to 20 are mixed up together and then a ticket is drawn at random. What is the probability that the ticket has a number which is a multiple of 3 or 7?

$\displaystyle n (S) = 20$

Let E be Number of events of getting a multiples of 3 or 7 on the drawn ticket.

$\displaystyle n(E) = 8 \text{ i.e. } \{ 3,6,9,12,15,18,7,14 \}$

$\displaystyle P (E) = \frac{n (E) }{ n (S)} = \frac{8 }{20} = \frac{2}{5}$

$\\$

Question 15: A bag contains 6 red, 4 white and 8 blue balls. If three balls are drawn at random, find the probability that one is red, one is white and one is blue.

If three balls are drawn at random,

$\displaystyle n (S) = ^{18}C_{3} = 3 \times 17 \times 16$

Let E be the event that one is red, one is white and one is blue

$\displaystyle n(E) = ^{6}C_{1} \times ^{4}C_{1} \times ^{8}C_{1} = 6 \times 4 \times 8$

$\displaystyle P (E) = \frac{n (E) }{ n (S)} = \frac{6 \times 4 \times 8 }{3 \times 17 \times 16} = \frac{4}{17}$

$\\$

Question 16: A bag contains 7 white,5 black and 4 red balls. If two balls are drawn at random, find the probability that (i) both the balls are white (ii) one ball is black and the other red (iii) both the balls are of the same color

A bag contains 7 white,5 black and 4 red balls. If two balls are drawn at random

Total number of possible outcomes is $\displaystyle = ^{16}C_{2} = 120$

$\displaystyle \text{Therefore, } n (S) = 120$

(i) both the balls are white

Let E be Number of events of getting a multiples of 3 or 7 on the drawn ticket.

$\displaystyle n(E) = 8 \text{ i.e. } \{ 3,6,9,12,15,18,7,14 \}$

$\displaystyle P (E) = \frac{n (E) }{ n (S)} = \frac{8 }{20} = \frac{2}{5}$

(ii) one ball is black and the other red

Let E be Number of events of getting a multiples of 3 or 7 on the drawn ticket.

$\displaystyle n(E) = 8 \text{ i.e. } \{ 3,6,9,12,15,18,7,14 \}$

$\displaystyle P (E) = \frac{n (E) }{ n (S)} = \frac{8 }{20} = \frac{2}{5}$

(iii) both the balls are of the same color

Let E be Number of events of getting a multiples of 3 or 7 on the drawn ticket.

$\displaystyle n(E) = 8 \text{ i.e. } \{ 3,6,9,12,15,18,7,14 \}$

$\displaystyle P (E) = \frac{n (E) }{ n (S)} = \frac{8 }{20} = \frac{2}{5}$

$\\$

Question 17: A bag contains 6 red, 4 white and 8 blue balls. If three balls are drawn at random, find the probability that (i) one is red and two are white (ii) two are blue and one is red (iii) one is red.

A bag contains 6 red,4 white and 8 blue balls. If three balls are drawn at random.

Total number of possible outcomes is $\displaystyle = ^{18}C_{3} = 816$

$\displaystyle \text{Therefore, } n (S) = 816$

(i) one is red and two are white

Let E be Number of events of one is red and two are white.

$\displaystyle n(E) = ^{6}C_{1} \times ^{4}C_{1} = 36$

$\displaystyle P (E) = \frac{n (E) }{ n (S)} = \frac{36 }{816} = \frac{3}{68}$

(ii) two are blue and one is red

Let E be Number of events of two are blue and one is red.

$\displaystyle n(E) = ^{8}C_{2} \times ^{6}C_{1} = 168$

$\displaystyle P (E) = \frac{n (E) }{ n (S)} = \frac{168 }{816} = \frac{7}{34}$

(iii) one is red

Let E be Number of events of one is red.

$\displaystyle n(E) = ^{6}C_{1} \times ^{4}C_{1} \times ^{8}C_{1} +^{6}C_{1} \times ^{4}C_{2} + ^{6}C_{1} \times ^{8}C_{2}= 396$

$\displaystyle P (E) = \frac{n (E) }{ n (S)} = \frac{396 }{816} = \frac{33}{68}$

$\\$

Question 18: Five cards are drawn from a pack of 52 cards. What is the chance that these 5 will contain: (i) just one ace      (ii) at least one ace?

Five cards are drawn from a pack of 52 cards.

Total number of possible outcomes is $\displaystyle = ^{52}C_{5} = 2598960$

$\displaystyle \text{Therefore, } n (S) = 2598960$

(i) just one ace

Let E be Number of events of just one ace.

$\displaystyle n(E) = ^{4}C_{1} \times ^{48}C_{4} = 778320$

$\displaystyle P (E) = \frac{n (E) }{ n (S)} = \frac{778320 }{2598960} = \frac{3243}{10829}$

(ii) at least one ace

Let E be Number of events of at least one ace.

$\displaystyle n(E) = ^{4}C_{1} \times ^{48}C_{4} + ^{4}C_{2} \times ^{48}C_{3} + ^{4}C_{3} \times ^{48}C_{4} + ^{4}C_{2} \times ^{48}C_{1} = 886656$

$\displaystyle P (E) = \frac{n (E) }{ n (S)} = \frac{886656 }{2598960} = \frac{18472}{54145}$

$\\$

Question 19: The face cards are removed from a full pack. Out of the remaining 40 cards,4 are drawn at random. What is the probability that they belong to different suits?

The face cards are removed from a full pack. Out of the remaining 40 cards,4 are drawn at random.

Total number of possible outcomes is $\displaystyle = ^{40}C_{4} = 91390$

$\displaystyle \text{Therefore, } n (S) = 91390$

Let E be the event that they belong to different suits

$\displaystyle n(E) = ^{10}C_{1} \times ^{10}C_{1} \times ^{10}C_{1} \times ^{10}C_{1} = 10000$

$\displaystyle P (E) = \frac{n (E) }{ n (S)} = \frac{10000}{91390} = \frac{1000}{9139}$

$\\$

Question 20: There are four men and six women on the city councils. If one council member is selected for a committee at random, how likely is that it is a women.

There are four men and six women on the city councils. If one council member is selected for a committee at random.

Total number of possible outcomes is $\displaystyle = ^{10}C_{1} = 10$

$\displaystyle \text{Therefore, } n (S) = 10$

Let E be the event that it is a women.

$\displaystyle n(E) = ^{6}C_{1} =6$

$\displaystyle P (E) = \frac{n (E) }{ n (S)} = \frac{6}{10} = \frac{3}{5}$

$\\$

Question 21: A box contains 100 bulbs, 20 of which are defective. 10 bulbs are selected for inspection. Find the probability that (i) all 10 are defective (ii) all 10 are good (iii) at least one is defective (iv) none is defective

A box contains 100 bulbs, 20 of which are defective. 10 bulbs are selected for inspection.

Total number of possible outcomes is $\displaystyle = ^{100}C_{10}$

$\displaystyle \text{Therefore, } n (S) = ^{100}C_{10}$

(i) all 10 are defective

Let E be the event when all 10 are defective.

$\displaystyle n(E) = ^{20}C_{10}$

$\displaystyle P (E) = \frac{n (E) }{ n (S)} = \frac{^{20}C_{10}}{^{100}C_{10}}$

(ii) all 10 are good

Let E be the event when all 10 are good.

$\displaystyle n(E) = ^{80}C_{10}$

$\displaystyle P (E) = \frac{n (E) }{ n (S)} = \frac{^{80}C_{10}}{^{100}C_{10}}$

(iii) at least one is defective

Let E be the event when at least one is defective.

Let E’ be the event when none of the bulds is defective.

$\displaystyle n(E') = ^{80}C_{10}$

$\displaystyle P (E') = \frac{n (E) }{ n (S)} = \frac{^{80}C_{10}}{^{100}C_{10}}$

$\text{Therefore } \displaystyle P (E) = 1 - P (E') = 1 - \frac{^{80}C_{10}}{^{100}C_{10}}$

(iv) none is defective

Let E be the event when all none is defective.

$\displaystyle n(E) = ^{80}C_{10}$

$\displaystyle P (E) = \frac{n (E) }{ n (S)} = \frac{^{80}C_{10}}{^{100}C_{10}}$

$\\$

Question 22: Find the probability that in a random arrangement of the letters of the word ‘SOCIAL’ vowels come together.

Number of letters in ‘SOCIAL’ = 6

Therefore total possible outcomes of arranging the alphabets are $\displaystyle 6!$

$\displaystyle n (S) = 6!$

Let E be the event that vowels come together

Number of vowels in SOCIAL is A, I, O

So, number of ways to arrange them where, (A, I, O) come together

$\displaystyle n(E) = 4! \times 3!$

$\displaystyle P (E) = \frac{n (E) }{ n (S)} = \frac{4! \times 3!}{6!} = \frac{1}{5}$

$\\$

Question 23:  The letters of the word ‘CLIFTON’ are placed at random in a row. What is the chance that two vowels come together?

Number of letters in ‘CLIFTON’ = 7

Therefore total possible outcomes of arranging the alphabets are $\displaystyle 7!$

$\displaystyle n (S) = 7!$

Let E be the event that vowels come together

Number of vowels in CLIFTON is I, O

So, number of ways to arrange them where, (I, O) come together

$\displaystyle n(E) = 6! \times 2!$

$\displaystyle P (E) = \frac{n (E) }{ n (S)} = \frac{6! \times 2!}{7!} = \frac{2}{7}$

$\\$

Question 24:  The letters of the word ‘FORTUNATES’ are arranged at random in a row. What is the chance that the two ‘T’ come together.

Number of letters in ‘FORTUNATES’ = 10

Therefore total possible outcomes of arranging the alphabets are $\displaystyle 10!$

$\displaystyle n (S) = 10!$

Let E be the event that T’s come together

So, number of ways to arrange them where, (T, T) come together

$\displaystyle n(E) = 9! \times 2!$

$\displaystyle P (E) = \frac{n (E) }{ n (S)} = \frac{9! \times 2!}{10!} = \frac{1}{5}$

$\\$

Question 25:  A committee of two persons is selected form two men and two women; What is the probability that the committee will have (i) no man ? (ii) one man? (iii) two men?

Total number of persons $\displaystyle = 2 + 2 = 4$

Out of the 4 persons, two can be selected in $\displaystyle ^{4}C_{2}$ ways.

$\displaystyle n(S) = ^{4}C_{2}$

(i) no man

Let E be the event that committee will have no man

$\displaystyle n(E) = ^{2}C_{2}$

$\displaystyle P (E) = \frac{n (E) }{ n (S)} = \frac{^{2}C_{2}}{^{4}C_{2}} = \frac{1}{\frac{4 \times 3}{2 \times 1}} = \frac{1}{6}$

(ii) one man

Let E be the event that committee will have one man

$\displaystyle n(E) = ^{2}C_{1} \times ^{2}C_{1}$

$\displaystyle P (E) = \frac{n (E) }{ n (S)} = \frac{^{2}C_{1} \times ^{2}C_{1}}{^{4}C_{2}} = \frac{2 \times 2}{\frac{4 \times 3}{2 \times 1}} = \frac{1}{3}$

(iii) two men

Let E be the event that committee will have two man

$\displaystyle n(E) = ^{2}C_{2}$

$\displaystyle P (E) = \frac{n (E) }{ n (S)} = \frac{^{2}C_{2}}{^{4}C_{2}} = \frac{1}{\frac{4 \times 3}{2 \times 1}} = \frac{1}{6}$

$\\$

Question 26:  If odds in favor of an event be 2:3, find the probability of occurrence of this event.

Odds in favor of an event be $\displaystyle 2:3$

Total number of possible outcomes $\displaystyle = 2k + 3k = 5k$

$\displaystyle n(S) = 5k$

Let E be the event that it occurs

$\displaystyle n(E) = 2k$

$\displaystyle P (E) = \frac{n (E) }{ n (S)} = \frac{2k}{5k} = \frac{2}{5}$

$\\$

Question 27: If odds against an event be 7:9, find the probability of non-occurrence of this event.

Odds against an event be $\displaystyle7:9$

Total number of possible outcomes $\displaystyle = 7k + 9k = 16k$

$\displaystyle n(S) = 16k$

Let E be the event that it does not occurs

$\displaystyle n(E) = 7k$

$\displaystyle P (E) = \frac{n (E) }{ n (S)} = \frac{7k}{16k} = \frac{7}{16}$

$\\$

Question 28: Two balls are drawn at random from a bag containing 2 white, 3 red, 5 green and 4 black balls, one by one without, replacement. Find the probability that both the balls are of different colors.

Two balls are drawn at random from a bag containing 2 white, 3 red, 5 green and 4 black balls, one by one without, replacement.

Total number of possible outcomes is $\displaystyle = ^{40}C_{4} = 91390$

$\displaystyle \text{Therefore, } n (S) = 91390$

Let E be the event that they belong to different suits

$\displaystyle n(E) = ^{10}C_{1} \times ^{10}C_{1} \times ^{10}C_{1} \times ^{10}C_{1} = 10000$

$\displaystyle P (E) = \frac{n (E) }{ n (S)} = \frac{10000}{91390} = \frac{1000}{9139}$

$\\$

Question 29: Two unbiased dice are thrown. Find the probability that:

(i) neither a doublet nor a total of 8 will appear

(ii) the sum of the numbers obtained on the two dice is neither a multiple of 2 nor a multiple of 3

Two unbiased dice are thrown.

Total number of possible outcomes is $\displaystyle = ^{6}C_{1} \times ^{6}C_{1} = 36$

$\displaystyle \text{Therefore, } n (S) = 36$

(i) neither a doublet nor a total of 8 will appear

Let E’ be the event that a doublet or a total of 8 occurs

$\displaystyle E'= 10 \text{ i.e. } \{ (1,1), (2,2), (3,3), (4,4) ,(5,5), (6,6), (2,6), (6,2) ,(3,5), (5,3) \}$

$\displaystyle n(E') = 10$

$\displaystyle P (E') = \frac{n (E') }{ n (S)} = \frac{10}{36} = \frac{5}{18}$

$\displaystyle P(E) = 1 - P(E') = 1 - \frac{5}{18} = \frac{13}{18}$

(ii) the sum of the numbers obtained on the two dice is neither a multiple of 2 nor a multiple of 3

Let E’ be the event that sum of number obtain on the dice is either a multiple of 2 or 3, that is total should be 2,3,4,6,8,9,10,12

$\displaystyle E'= 24 \text{ i.e. } \{ (1,1) ,(1,2), (2,1), (1,3), (2,2), (3,1), \\ (1,5), (2,4), (3,3), (4,2), (5,1) ,(2,6), (3,5), (4,4) ,(5,3), (6,2), (3,6), \\ (4,5), (5,4), (6,3), (4,6), (5,5), (6,4) ,(6,6) \}$

$\displaystyle n(E') = 24$

$\displaystyle P (E') = \frac{n (E') }{ n (S)} = \frac{24}{36} = \frac{2}{3}$

$\displaystyle P(E) = 1 - P(E') = 1 - \frac{2}{3} = \frac{1}{3}$

$\\$

Question 30: A bag contains 8 red, 3 white and 9 blue balls. If three balls are drawn at random, determine the probability that (i) all the three balls are blue balls (ii) all the balls are of different colors.

A bag contains 8 red, 3 white and 9 blue balls and three balls are drawn at random.

Total number of possible outcomes is $\displaystyle = ^{20}C_{3} = 1140$

$\displaystyle \text{Therefore, } n (S) = 1140$

(i) all the three balls are blue balls

Let E be the event that all the three balls are blue balls

$\displaystyle n(E) = ^{9}C_{3} = 84$

$\displaystyle P (E) = \frac{n (E) }{ n (S)} = \frac{84}{1140} = \frac{7}{95}$

(ii) all the balls are of different colors

Let E be the event that all the balls are of different colors

$\displaystyle n(E) = ^{8}C_{1} \times ^{3}C_{1} \times ^{9}C_{1} =8 \times 3 \times 9 = 216$

$\displaystyle P (E) = \frac{n (E) }{ n (S)} = \frac{216}{1140} = \frac{18}{95}$

$\\$

Question 31: A bag contains 5 red, 5 white and 7 black balls. Two balls are drawn at random. What is the probability that both balls are red or both are black?

A bag contains 5 red, 5 white and 7 black balls. Two balls are drawn at random.

Total number of possible outcomes is $\displaystyle = ^{18}C_{2} = 153$

$\displaystyle \text{Therefore, } n (S) = 153$

Let E be the event that both balls are red or both are black

$\displaystyle n(E) = ^{5}C_{2} + ^{7}C_{2} = 31$

$\displaystyle P (E) = \frac{n (E) }{ n (S)} = \frac{31}{153}$

$\\$

Question 32: If a letter is chosen at random from the English alphabet, find the probability that the letter is (i) a vowel (ii) a consonant

A letter is chosen at random from the English alphabet.

Total number of possible outcomes is $\displaystyle = ^{26}C_{1} = 26$

$\displaystyle \text{Therefore, } n (S) = 26$

(i) a vowel

Let E be the event that the letter is a vowel

Favorable outcomes are $\displaystyle a, e, i, o, u$

Total number of favorable outcomes = 5

$\displaystyle P (E) = \frac{n (E) }{ n (S)} = \frac{5}{26}$

(ii) a consonant

Let E be the event that the letter is a consonant ( not a vowel)

Total number of favorable outcomes $\displaystyle = 26-5 = 21$

$\displaystyle P (E) = \frac{n (E) }{ n (S)} = \frac{21}{26}$

$\\$

Question 33: In a lottery, a person chooses six different numbers at random from 1 to 20, and if these six numbers match with six numbers already fixed by the lottery committee, he wins the prize. What is the probability of winning the prize in the game?

A person chooses six different numbers at random from 1 to 20, and if these six numbers match with six numbers already fixed by the lottery committee, he wins the prize.

Total number of possible outcomes is $\displaystyle = ^{20}C_{6} = 38760$

$\displaystyle \text{Therefore, } n (S) = 38760$

Let E be the event that all six numbers match with the given number (as winning number is fixed).

$\displaystyle n(E) = 1$

$\displaystyle P (E) = \frac{n (E) }{ n (S)} = \frac{1}{38760}$

$\\$

Question 34: 20 cards are numbered from 1 to 20. One card is drawn at random. What is the probability that the number on the cards is: (i) a multiple of 4? (ii) not a multiple of 4? (iii) odd? (iv) greater than 12? (v) divisible by 5? (vi) not a multiple of 6?

20 cards are numbered from 1 to 20. One card is drawn at random.

Total number of possible outcomes is $\displaystyle = ^{20}C_{1} = 20$

$\displaystyle \text{Therefore, } n (S) = 20$

(i) a multiple of 4

Let E be the event that the number on the drawn card is a multiple of 4

$\displaystyle n(E) = 5 \text{ i.e. } \{ 4, 8, 12, 16, 20 \}$

$\displaystyle P (E) = \frac{n (E) }{ n (S)} = \frac{5 }{20} = \frac{1}{4}$

(ii) not a multiple of 4

Let E the event that the number on the drawn card is not a multiple of 4

$\displaystyle n(E') = 5 \text{ i.e. } \{ 4, 8, 12, 16, 20 \}$

$\displaystyle P (E') = \frac{n (E') }{ n (S)} = \frac{5 }{20} = \frac{1}{4}$

$\displaystyle P(E) = 1 - P(E') = 1 - \frac{1}{4} = \frac{3}{4}$

(iii) odd

Let E be the event that the number on the drawn card is odd

$\displaystyle n(E) = 10 \text{ i.e. } \{ 1,3,5,7,9,11,13,15,17,19 \}$

$\displaystyle P (E) = \frac{n (E) }{ n (S)} = \frac{10 }{20} = \frac{1}{2}$

(iv) greater than 12

Let E be the event that the number on the drawn card is greater than 12

$\displaystyle n(E) = 8 \text{ i.e. } \{ 13, 14, 15, 16, 17, 18, 19, 20 \}$

$\displaystyle P (E) = \frac{n (E) }{ n (S)} = \frac{8 }{20} = \frac{2}{5}$

(v) divisible by 5

Let E be the event that the number on the drawn card is divisible by 5

$\displaystyle n(E) = 4 \text{ i.e. } \{ 5, 10, 15, 20 \}$

$\displaystyle P (E) = \frac{n (E) }{ n (S)} = \frac{4 }{20} = \frac{1}{5}$

(vi) not a multiple of 6

Let E be the event that the number on the drawn card is divisible by 6

$\displaystyle n(E') = 3 \text{ i.e. } \{ 6, 12, 18 \}$

$\displaystyle P (E') = \frac{n (E) }{ n (S)} = \frac{3 }{20}$

$\displaystyle P(E) = 1 - P(E') = 1 - \frac{3 }{20} = \frac{17 }{20}$

$\\$

Question 35: Two dice are thrown. Find the odds in favor of getting the sum (i) 4 (ii) 5 (iii) what are the odds against getting the sum 6?

Total number of possible outcomes is $\displaystyle = 6 \times 6 = 36$

$\displaystyle \text{Therefore, } n (S) = 36$

(i) sum is 4

Let E be the event that the sum is 4

$\displaystyle n(E) = 3 \text{ i.e. } \{ (1,3), (2, 2), (3, 1) \}$

$\displaystyle P (E) = \frac{n (E) }{ n (S)} = \frac{3 }{36} = \frac{1}{11}$

(ii) sum is 5

Let E be the event that the sum is 5

$\displaystyle n(E) = 4 \text{ i.e. } \{ (1,4), (2, 3), (3, 2), (4, 1) \}$

$\displaystyle P (E) = \frac{n (E) }{ n (S)} = \frac{4 }{36} = \frac{1}{9}$

(iii) what are the odds against getting the sum 6?

Let E’ be the event that the sum is 6

$\displaystyle n(E') = 5 \text{ i.e. } \{ (1,5), (2, 4), (3, 3), (4, 2), (5, 1) \}$

$\displaystyle P (E') = \frac{n (E) }{ n (S)} = \frac{5 }{36}$

$\displaystyle P(E) = 1 - P(E') = 1 - \frac{5 }{36} = \frac{31 }{36}$

$\\$

Question 36: What are the odds in favor of getting a spade if a card is drawn from a well-shuffled deck of cards? What are the odds in favor of getting a king?

A card is drawn from a well-shuffled deck of cards.

Total number of possible outcomes is $\displaystyle = ^{52}C_{1} = 52$

$\displaystyle \text{Therefore, } n (S) = 52$

(i) Let E be the event that we get a spade

$\displaystyle n(E) = 13 \text{ i.e. } \{ 13 \text{ Spade cards } \}$

$\displaystyle P (E) = \frac{n (E) }{ n (S)} = \frac{13 }{52} = \frac{1}{4}$

Therefore, probability of event E’ is

$\displaystyle P(E') = 1 - P(E) = 1 - \frac{1 }{4} = \frac{3 }{4}$

$\displaystyle \text{Odds in favour of getting a spade is } \frac{P(E)}{P(E')} = \frac{\frac{1}{4}}{\frac{3 }{4}} =1:3$

(ii) Let E be the event that we get a king

$\displaystyle n(E) = 4 \text{ i.e. } \{ 4 \text{ Kings } \}$

$\displaystyle P (E) = \frac{n (E) }{ n (S)} = \frac{4 }{52} = \frac{1}{13}$

Therefore, probability of event E’ is

$\displaystyle P(E') = 1 - P(E) = 1 - \frac{1 }{13} = \frac{12 }{13}$

$\displaystyle \text{Odds in favour of getting a king is } \frac{P(E)}{P(E')} = \frac{\frac{1}{13}}{\frac{12 }{13}} =1:12$

$\\$

Question 37: A box contains 10 red marbles, 20 blue marbles and 30 green marbles. 5 marbles are drawn at random. From the box, what is the probability that (i) all are blue? (ii) at least one is green?

A box contains 10 red marbles, 20 blue marbles and 30 green marbles. 5 marbles are drawn at random.

Total number of possible outcomes is $\displaystyle = ^{60}C_{5}$

$\displaystyle \text{Therefore, } n (S) = ^{60}C_{5}$

(i) all are blue?

Let E be the event that we get all blue

5 blue marbles out of 20 can be drawn in $\displaystyle ^{20}C_{5}$ ways

$\displaystyle n(E) = ^{20}C_{5}$

$\displaystyle P (E) = \frac{n (E) }{ n (S)} = \frac{^{20}C_{5} }{^{60}C_{5}} = \frac{20 \times 19 \times 18 \times 17 \times 16}{60\times 59\times 58\times 57 \times 56} = \frac{ 2 \times 17 }{59 \times 29 \times 7} = \frac{34}{11977}$

(ii) at least one is green

Let E’ be the event that no green marble is chosen

5 non-green marbles out of 30 can be drawn in $\displaystyle ^{30}C_{5}$ ways

$\displaystyle n(E') = ^{30}C_{5}$

$\displaystyle P (E') = \frac{n (E) }{ n (S)} = \frac{^{30}C_{5} }{^{60}C_{5}}$

$\displaystyle P(E) = 1 - P(E') = 1 - \frac{^{30}C_{5} }{^{60}C_{5}} = 1 - \frac{30 \times 39 \times 38 \times 37 \times 36 }{60 \times 59 \times 58 \times 57 \times 56 } \\ \\ \\ = 1 - \frac{9 \times 13}{4 \times 59 \times 19} = 1- \frac{117}{4484} = \frac{4367}{4484}$

$\\$

Question 38: A box contains 6 red marbles numbered 1 through 6 and 4 white marbles numbered from 12 through 15. Find the probability that a marble drawn is (i) white (ii) white and odd numbered (iii) even numbered (iv) red or even numbered.

A box contains 6 red marbles numbered 1 through 6 and 4 white marbles numbered from 12 through 15. One marble is drawn.

Total number of possible outcomes is $\displaystyle = ^{10}C_{1} = 10$

$\displaystyle \text{Therefore, } n (S) = 10$

(i) white

Let E be the event of getting white marble

$\displaystyle n(E) = ^{4}C_{1} = 4$

$\displaystyle P (E) = \frac{n (E) }{ n (S)} = \frac{4 }{10} = \frac{2}{5}$

(ii) white and odd numbered

Let E be the event of getting white and odd numbered marble

$\displaystyle n(E) = 2 \text{ i.e. } \{ 13, 15 \}$

$\displaystyle P (E) = \frac{n (E) }{ n (S)} = \frac{2 }{10} = \frac{1}{5}$

(iii) even numbered

Let E be the event of getting even numbered marble

$\displaystyle n(E) = 5 \text{ i.e. } \{ 2, 4, 6, 12, 14 \}$

$\displaystyle P (E) = \frac{n (E) }{ n (S)} = \frac{5 }{10} = \frac{1}{2}$

(iv) red or even numbered

Let E be the event of getting red or even numbered marble

$\displaystyle n(E) = 8 \text{ i.e. } \{1, 2, 3, 4, 5, 6, 12, 14\}$

$\displaystyle P (E) = \frac{n (E) }{ n (S)} = \frac{8 }{10} = \frac{4}{5}$

$\\$

Question 39: A class consists of 10 boys and 8 girls. 3 students are selected at random. What is the probability that the selected group has (i) all boys? (ii) all girls? (iii) 1 boy and 2 girls? (iv) at least one girl? (v) at most one girl?

A class consists of 10 boys and 8 girls. 3 students are selected at random.

Total number of possible outcomes is $\displaystyle = ^{18}C_{3} = 816$

$\displaystyle \text{Therefore, } n (S) = 816$

(i) all boys

Let E be the event of getting all boys

$\displaystyle n(E) = ^{10}C_{3} =120$

$\displaystyle P (E) = \frac{n (E) }{ n (S)} = \frac{120 }{816} = \frac{5}{34}$

(ii) all girls

Let E be the event of getting all girls

$\displaystyle n(E) = ^{8}C_{3} =56$

$\displaystyle P (E) = \frac{n (E) }{ n (S)} = \frac{56 }{816} = \frac{7}{102}$

(iii) 1 boy and 2 girls

Let E be the event of getting 1 boy and 2 girls

$\displaystyle n(E) = ^{10}C_{1} \times ^{8}C_{2} =280$

$\displaystyle P (E) = \frac{n (E) }{ n (S)} = \frac{280 }{816} = \frac{35}{102}$

(iv) at least one girl

Let E be the event of getting at least one girl

$\displaystyle n(E) = ^{10}C_{2} \times ^{8}C_{1} + ^{10}C_{1} \times ^{8}C_{2} + ^{10}C_{0} \times ^{8}C_{3}=696$

$\displaystyle P (E) = \frac{n (E) }{ n (S)} = \frac{696 }{816} = \frac{29}{34}$

(v) at most one girl

Let E be the event of getting atmost one girl

$\displaystyle n(E) = ^{10}C_{2} \times ^{8}C_{1} + ^{10}C_{3} \times ^{8}C_{0} =480$

$\displaystyle P (E) = \frac{n (E) }{ n (S)} = \frac{480 }{816} = \frac{10}{17}$

$\\$

Question 40: Five cards are drawn from a well-shuffled pack of 52 cards. Find the probability that all the
five cards are hearts.

Five cards are drawn from a well-shuffled pack of 52 cards.

Total number of possible outcomes is $\displaystyle = ^{52}C_{5} =2598960$

$\displaystyle \text{Therefore, } n (S) = 2598960$

Let E be the event that all the five cards are hearts.

$\displaystyle n(E) = ^{13}C_{5} = 1287$

$\displaystyle P (E) = \frac{n (E) }{ n (S)} = \frac{1287 }{2598960} = \frac{33}{66640}$

$\\$

Question 41: A bag contains tickets numbered from 1 to 20. Two tickets are drawn. Find the probability that (i) both the tickets have prime numbers on them (ii) on one there is a prime number and on the other there is a multiple of 4.

A bag contains tickets numbered from 1 to 20. Two tickets are drawn.

Total number of possible outcomes is $\displaystyle = ^{20}C_{2} =190$

$\displaystyle \text{Therefore, } n (S) = 190$

(i) both the tickets have prime numbers on them

Let E be the event that both the tickets have prime numbers on them.

$\displaystyle E= \{ 2,3,5,7,11,13,17,19 \}$

$\displaystyle n(E) = ^{8}C_{2} = 28$

$\displaystyle P (E) = \frac{n (E) }{ n (S)} = \frac{28 }{190} = \frac{28}{190}$

(ii) one there is a prime number and on the other there is a multiple of 4

Let E be the event that one there is a prime number and on the other there is a multiple of 4.

$\displaystyle E= \{ 2,3,5,7,11,13,17,19 \}$ for prime number

$\displaystyle E= \{ 4,8,12,16,20 \}$ for multiple of 4

$\displaystyle n(E) = ^{8}C_{1} \times ^{5}C_{1}= 40$

$\displaystyle P (E) = \frac{n (E) }{ n (S)} = \frac{40 }{190} = \frac{4}{19}$

$\\$

Question 42: An urn contains 7 white, 5 black and 3 red balls. Two balls are drawn at random. Find the probability that (i) both the balls are red (ii) one ball is red and the other is black (iii) one ball is white.

An urn contains 7 white, 5 black and 3 red balls. Two balls are drawn at random.

Total number of possible outcomes is $\displaystyle = ^{18}C_{2} =105$

$\displaystyle \text{Therefore, } n (S) = 105$

(i) both the balls are red

Let E be the event that both the balls are red

$\displaystyle n(E) = ^{3}C_{2} = 3$

$\displaystyle P (E) = \frac{n (E) }{ n (S)} = \frac{3 }{105} = \frac{1}{35}$

(ii) one ball is red and the other is black

Let E be the event that one ball is red and the other is black

$\displaystyle n(E) = ^{3}C_{1} \times ^{5}C_{1} = 15$

$\displaystyle P (E) = \frac{n (E) }{ n (S)} = \frac{15}{105} = \frac{1}{7}$

(iii) one ball is white

Let E be the event that one ball is white

$\displaystyle n(E) = ^{8}C_{1} \times ^{7}C_{1} = 56$

$\displaystyle P (E) = \frac{n (E) }{ n (S)} = \frac{56}{105} = \frac{8}{15}$

$\\$

Question 43: A and B throw a pair of dice. If A throws 9, find B’s chance of throwing a higher number.

Two dices are thrown once.

Total number of possible outcomes is $\displaystyle 6^2 = 36$

$\displaystyle \text{Therefore, } n (S) = 36$

Let E be the event of B throwing a no. higher than 9

No. of favorable outcomes $\displaystyle = 6 \{ (4, 6), (5, 5), (6, 4), (5, 6) ,(6, 5), (6, 6) \}$

$\displaystyle n(E) = 6$

$\displaystyle P (E) = \frac{n (E) }{ n (S)} = \frac{6}{36} = \frac{1}{6}$

$\\$

Question 44: In a hand at Whist, what is the probability that four kings are held by a specified player?

We know that 13 cards are delivered to a hand at whist.

Total number of possible outcomes is $\displaystyle ^{52}C_{13}$

$\displaystyle \text{Therefore, } n (S) = ^{52}C_{13}$

Since 4 kings are held by a specified player, 9 more cards are to be delivered to him from the remaining 48 cards

No of favorable ways $\displaystyle = ^{48}C_{9}$

$\displaystyle P (E) = \frac{n (E) }{ n (S)} = \frac{^{48}C_{9}}{^{52}C_{13}} = \frac{13 \times 12 \times 11 \times 10}{52 \times 51 \times 50 \times 49 } = \frac{11}{4165}$

$\\$

Question 45: Find the probability that in a random arrangement of the letters of the word ‘UNIVERSITY’, the two I’s do not come together.

Let S be the sample space.

Then, $\displaystyle n(S) =$ Total number of ways in which the letters of the word  $\displaystyle \text{UNIVERSITY can be arranged} = \frac{10!}{2!}$

Let E be the event of arranging the letters of the word ‘UNIVERSITY’ such that 2 I’s remain together.

Number of ways where the two Is are always together $\displaystyle = 9!$

$\displaystyle \text{Therefore the number of ways where the 2Is are not together } = \frac{10!}{2!} - 9!$

$\displaystyle \text{Therefore the probability } = \frac{\frac{10!}{2!} - 9!}{\frac{10!}{2!}} = \frac{4}{5}$

$\\$


text:
# How to apply oxidation numbers in the reaction to form hydrogen peroxide from a superoxide?

In the following reaction: $$\ce{2O2- (aq) + 2H+ (aq) <=> H2O2 (aq) + O2(g)}$$

I see where some oxygen atoms have lost electrons, but I do not see what has gained electrons.
On the reactant side, the superoxide has a $$-1$$ charge per oxygen atom. On the product side, half the oxygen atoms now have a $$0$$ charge, so they lost an electron each (right?). For it to be a true redox reaction, does not a species/atom have to gain electron(s)?

On the reactant side, the superoxide has a $$-1$$ charge per oxygen atom.

No, the superoxide has a $$-\frac{1}{2}$$ charge per oxygen atom.

The reaction is better thought of as two separate reactions:

\begin{align} \ce{O2- &<=> O2 + e-}\tag1\\ \ce{O2- + 2H+ + e- &<=> H2O2}\tag2 \end{align}

This demarcation is particularly relevant in a biological context, because there is a class of enzymes Superoxide reductase (SOR) that only catalyze the reaction forming $$\ce{H2O2}$$ and another class of enzymes Superoxide dismutase (SOD) that catalyzes both.

Here, oxygen is the element which is simultaneously undergoing oxidation (equation (1)) and reduction (equation (2)). This is termed a disproportionation reaction.


text:
## Archive for June, 2010

### Check your stereo speaker setup

Wednesday, June 30th, 2010

When ever I set up the laptop with those little computer speakers I have to check the stereo setup: so I’m sure the right speaker is on the right and the left on the left. On the Mac I do this by going to System Preferences > Sound then switching the sound all the way to the right and flipping the volume up and down to make little bleeps. Then the same for the left.

Finally here’s a definitive webapp solution:

### Vi(m) tip #9: Copy, Cut and Paste into Mac OS X clipboard

Wednesday, June 30th, 2010

I can get a lot done in vim without ever having to use the mouse or really exit “vim world”. But the one thing that I keep falling back on is copy and pasting from and to other applications. Here’s a way to do these
operations without resorting to the mouse or foreign keystrokes. Issue these in command mode:

### Cut line under cursor

:.!pbcopy

### Copy line under cursor

:.!pbcopy|pbpaste

### Paste as line beneath cursor

:r !pbpaste

You can also use this if you’ve made a selection in Visual Mode:

### Cut current selection

(warning: this grabs the whole lines of any lines within selection)

:'<,'>!pbcopy

### Copy current selection

(warning: this grabs the whole lines of any lines within selection)

:'<,'>!pbcopy|pbpaste

Note: It seems for that one you should just select the first letter of each line. Not the entire block, then it just cuts…

Update: It’s much simpler:

"*yy

Yanks the current line to the clipboard. Similarly,

"*dd

cuts the current line and

"*p

pastes the clipboard below the current line.

### OpenGL texture map all white

Monday, June 28th, 2010

After a long battle today, I found a bug that was making all my textures render white. Here’s pseudo code of what I had:

### Bugged code

GLwidget.h

class GLwidget : QGLWidget {
GLWidget();
...
void initializeGL();
void paintGL();
...
}

GLwidget.cpp

GLWidget::GLWidget(){
// code that generates texture
makeMyTexture(...);
// opengl calls that setup texture in GL land
glGenTextures(1, &texName);
glBindTexture(GL_TEXTURE_2D, texName);
glTexParameteri(GL_TEXTURE_2D, GL_TEXTURE_WRAP_S, GL_REPEAT);
...
glTexImage2D(GL_TEXTURE_2D, ...) ;
}
...
void GLWidget ::initializeGL(){
// opengl initialization code
}
void GLWidget :: paintGL(){
// code that actually uses the texture
glEnable(GL_TEXTURE_2D)
...
glDisable(GL_TEXTURE_2D)
}

My GLWidget class inherited from a class that called initializeGL and paintGL at the appropriate times. The only problem was that I was building and setting my texture in the class constructor before initializeGL() was ever called. Then when it was, it wiped out my texture.

The correct order should be:

### Correct

GLwidget.cpp

GLWidget::GLWidget(){
...
}
void GLWidget ::initializeGL(){
// opengl initialization code
...
// code that generates texture
makeMyTexture(...);
// opengl calls that setup texture in GL land
glGenTextures(1, &texName);
glBindTexture(GL_TEXTURE_2D, texName);
glTexParameteri(GL_TEXTURE_2D, GL_TEXTURE_WRAP_S, GL_REPEAT);
...
glTexImage2D(GL_TEXTURE_2D, ...) ;
}
void GLWidget :: paintGL(){
// code that actually uses the texture
glEnable(GL_TEXTURE_2D)
...
glDisable(GL_TEXTURE_2D)
}

Initialize your texture after you have initialized opengl.

### X11 libraries not found on Mac OS X 10.6

Monday, June 28th, 2010

I never had this trouble on my mac os x 10.5 machine, but now on 10.6 I’m getting errors when including X11 libraries. When I try to compile using -lX11 or -lXext I get complaints from the linker like

or

To fix this you could modify your environment paths, or add the following to your compile flags:

-L/usr/X11/lib

I do that and everything compiles and links fine.

### One time password for single use guest key

Saturday, June 26th, 2010

What’s the online equivalent to leaving your house key under the doormat?

A number of times I have been caught in a situation where I’m not at a computer and need to get something off of my email. I call a friend, entrust them with my password, retrieve the information and proceed to worry that I have (1) allowed my friend to (un)intentionally view something private of mine or (2) at least put my friend in the unfair situation where he needs to convince himself not to snoop. Whether he ever would or should is not the point, the situation of my giving him such a powerful position is bad enough.

Here’s a solution:
When I sign up for a new email account, I should give a normal password and a “one time password”. I don’t tell anyone my OTP. But then if I ever need to let a friend into my account I give them this OTP that works just once.

The email provider could come up with all sorts of different schemes for how the “just once” is defined. By time: logs off automatically after 60 seconds. By activity: logs off after a n searches. Or just limit access for a single session: only able to view certain emails or of a certain age…blah blah blah.

It could be as secure as you wanted to be. The point is that the current situation isn’t secure at all: you have to give out your entire password. I guess the reason giving out your password is not the same as leaving a key under your doormat is that your key is a single object and it doesn’t unlock every door in your house much less provide you with a simple way to search for valuable items.

### Barycentric versus voronoi regional area in mass matrices

Friday, June 25th, 2010

In a recent project, we were minimizing an energy over a surface approximated by a triangle mesh. Our finite element method approach reduced our energy to solving a system of linear equations with coefficients from a stiffness matrix and mass matrix. In the lumped version of a mesh’s mass matrix, diagonal entries correspond to areas on the surface around each vertex. You want all the areas to sum to the total area of the surface and you would like the area for each vertex to some how reflect the influence of its incident triangles.

### Barycentric

One natural choice is a barycentric area. Take the barycenter (centroid, center of mass), which lies at the intersection of medians. Using those medians divide the triangle into three quadrilaterals each corresponding its incident point:

These quadrilaterals are by definition of equal area. The barycentric mass matrix entry for vertex i is the sum of its corresponding quadrilaterals from all incident triangles, or more simply: one-third of the total area of all neighboring triangles. Notice that the contribution of a triangle to its vertices is the same for all three, regardless of there placement around the triangle.

### Voronoi

Another, also natural, choice is voronoi area. Take the circumcenter, which lies at the center of the circumscribing circle of the triangle. It follows quickly that the circumcenter is equidistant from all vertices on the triangle. Connect this point to the midpoints along the edges of the triangle, again forming three quadrilaterals.

Note that this time the quadrilaterals are not equal. They are not even guaranteed to be positive. For obtuse triangles, the acute angled vertices will receive negative areas (assuming you demand their sums equal the area of the triangle). Here is a plot of voronoi areas as the angle of an isosceles triangle grows from 0 (acute) to π (obtuse). The x axis is the angle of the isosceles triangle and the y axis is area. The green line shows the area of corresponding to the vertex at the angle that’s growing, the blue corresponds to the other two areas.

Notice that at π/2 we might expect the other two areas to cross into negative but actually this happens later. So voronoi areas won’t produce negative coefficients until fairly obtuse.

Again, the voronoi mass matrix entry for vertex i is the sum of its corresponding quadrilaterals from all incident triangles. Since areas can be negative, this means negative entries could end up in the voronoi mass matrix.

### Connectivity dependence

The Voronoi area of a vertex is called so because the area represents its corresponding voronoi region: all the points on the surface for which this vertex is the closest vertex (assuming the mesh has “nice” triangles: non-obtuse). In this way the voronoi area depends only on the vertices positioning on the surface not the connectivity of the mesh.

The barycentric area, while having the advantage of always being positive, is heavily connectivity dependent as a vertex gets one-third of the area of any connected triangle.

For a simple example, here’re two meshes. One with regular positions and regular connectivity:

And one with regular positions and the same regular connectivity but a single edge at each highlighted vertex has been flipped:

And here are the barycentric (left) and voronoi (right) areas for the highlighted vertices on the regular connectivity mesh.

Now, notice how when the edge is flipped the barycentric area changes drastically, and actually changes for all vertices incident on the flipped edge. The voronoi however stays the same for this vertex and all vertices.

### Hybrid approach to vonoroi

To handle the problem of negative voronoi areas for some obtuse triangles, there exists a hybrid approach. For acute triangles, use voronoi as is. For obtuse triangles snap the circumcenter to the midpoint along the edge opposite the large angle and use the corresponding quads (I described this in an earlier post). Notice for a right angle triangle the circumcenter corresponds with this midpoint, so as a triangle becomes obtuse there isn’t a jump in area distribution. A keen eye will notice that I am doing this in the voronoi area picture above for the obtuse triangle.

### Euclidean (floored) modulo (mod, %) operator for C, C++

Thursday, June 24th, 2010

Once again, I was gotcha-ed by the sign of the mod operator in C++. I did -1%n, hoping to get n-1 but instead got -1.
Here’s a little C/C++ function (you could even macro/inline it) to give the “true” (Euclidean) modulo:

static int mod(int a, int base){
return ((a % base) + base) % base;
}

Update: So the above runs about twice as slow as just issuing % once, which is to be expected since % is called twice (guess + is pretty cheap in comparison). If half the time a will be positive and half the time negative then a average speed up of 1.5x over the above code can be achieved with the following:

static int mod(int a, int base){
return (a < 0 ? ((a % base) + base) % base : a % base);
}

But it is still slower if you already know for sure that a is not negative.

Update: It's worth noting, as it was noted in the comments, that what I describe above is only correct if the base is positive. If the base is negative than euclidean modulo allows a negative result and must be defined accordingly.

### Vi(m) tip #8: scroll cursor to top, center or bottom of screen

Wednesday, June 23rd, 2010

In command mode you can do some handy quick scrolling.

Scroll so that line with cursor is in the center of the screen:

z.

Scroll so that line with cursor is at the top of the screen:

zt

Scroll so that line with cursor is at the bottom of the screen:

zb

source

### Brightness Menulet (revised)

Monday, June 21st, 2010

I’ve taken another swing at making a brightness menulet like the volume menulet. My old version could only toggle brightness (because of UI problems). But the new version is a full on slider!

Here’s the xcode project and intel/ppc builds for at least mac os x 10.5:
Brightness Menulet Xcode Project

Note: This menulet (and the original brightness code) does not work with 10.6 on my Mac Cinema HD display, but it does work with 10.6 on my girlfriend’s MacBook.

### Split (large) files in half using bash script

Monday, June 21st, 2010

Here’s a simple bash script that splits a given file into two equal (up to a byte) halves. It uses split which has different syntax between the GNU and BSD implementations, so see the comments for switching between the two.
I tested this on a 6.4 GB dmg file and I was able to reverse the split with cat just fine. I diffed the re-concatenated file and the original and they were byte-wise equal.

Save this in a file called half.sh:

#!/bin/bash
#
# usage: half foo
#
# This will create foo.aa and foo.ab, where foo.aa is the first half(+1) of foo
# and foo.ab is the second half. Split by bytes not lines.
#
#
# To reverse just issue:
# cat foo.aa foo.ab > foo
#

# get file size
# GNU:
#size=stat -c "%s" "$1" # BSD: size=stat -f "%z" "$1"

# divide by 2 and round up
half_size=echo "($size+1)/2" | bc split -b$half_size "$1" "$1."


text:
# Gel filtration separation of cellulose, starch, oxytocin and palmitic acid

I am studying biochemistry and I have been asked by my teacher to find a procedure for separating cellulose, starch, the peptide oxytocin (9 a.a) and palmitic acid ($\ce{C16}$) using gel filtration chromatography. I have some ideas about using Fehling's test to tell the cellulose from the starch and the peptide since they all react with the $\ce{Cu^{2+}}$ in the solution. And I'm thinking that the palmitic acid will not dissolve in the solution and can therefore be separated from the others.

However, I have a problem with how the gel filtration would work. Obviously cellulose and starch are much bigger than palmitic acid and oxytocin but, how could I possibly separate cellulose from starch using gel filtration??

Gel filtration: https://en.wikipedia.org/wiki/Size-exclusion_chromatography

• Are you sure you have to use gel filtration for this? Cellulose might be the right size for gel filtration, but starch is normally not soluble and the two smaller molecules, oxytocin and palmitic acid will probably elute in the void volume. – VonBeche Dec 8 '16 at 17:03


text:
## College Algebra 7th Edition

a) $x=2$ b)$x=\frac{5}{2}$ c)$x=-2$ d)$x=-1$
a) Add $1$ to both sides and divide. $3x-1=5$ $3x=6$ $x=2$ b) Subtract $3$ from both sides and divide. $2x+3=8$ $2x=5$ $x=\frac{5}{2}$ c) Subtract 5x from both sides and divide. $2x=5x+6$ $-3x=6$ $x=-2$ d) Add $4x-11$ to both sides and divide. $x+11=6-4x$ $5x=-5$ $x=-1$