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# What is the point of not considering the OP's upvote for Unsung Hero badge?

I have been reading some questions about the Unsung Hero badge (applies to Tenacious too), and it seems this badge is a little controversial (questions arguing if it is a good thing to pursue the badge, the gamification that surrounds it e.g. the guy asks to the OP to not upvote the question, etc).

What is the point of not considering the +1 (upvote) that generally comes from the OP?

I mean, many users upvote answers when accepting them. Some users do not upvote.

If the badge's goal is to reward users who answer questions that do not get much attention, what is the difference under the point of view of getting attention when a question is accepted and has zero score, or is accepted and have 1 upvote?

Ok, the sole upvote does not always come from the OP, but surely the great majority of cases it does.

Tenacious (silver): Maximum of +1 score accepted answers: more than 5 and 20% of total;

Unsung Hero (gold): Maximum of +1 score accepted answers: more than 10 and 25% of total.

• If the upvote from the OP is considered, then it is possible to be a slightly more-sung hero. – Aza Jul 31 '13 at 0:57
• I can't believe there are over 5,000 users with the Unsung Hero badge on SO. I expected a much smaller number. We have ZERO on dba.SE. – Aaron Bertrand Jul 31 '13 at 1:10

I agree with the sentiment of this discussion. For a while now I've been thinking that the advice to "upvote and accept" runs counter to Tenacious and Unsung Hero.

One possibility would be to ignore only an OP upvote when awarding these badges. This would correctly disqualify answers upvoted by some random party, but would leak some information about the identity of the sole upvoter (the OP), which may be a bad thing. (This would also make it slightly more difficult to correctly word the badge descriptions; how about

(gold) Unsung Hero: At least 10 accepted answers, of which 25% have not been upvoted (except by the asker).

?)

# Response to AsheeshR and jmac

The above users argue that:

2. If measurement were impossible, people would be less motivated to get the badge.
3. If measurement were possible, lone OP upvotes would be de-anonymised, which would be bad.

## On the tracking side

I don't have much data either, but I don't believe that the de-anonymisation would be much of a net negative. It is unlikely that all that many of the questions counting towards your badge were asked by the same person, making it difficult to collect aggregate statistics; also "otherwise-zero-score-answer-upvote rate" probably isn't a priority to stick on a usercard like accept rate was (wrongly) judged to be, and it's difficult to determine by other means.

## On the non-tracking side

Is tracking progress so critical? I was awarded Tenacious just as a side-effect of answering questions and helping people. If anything, not knowing how close (or far away) they are from Unsung Hero will just lead people to try and post more answers in lonely sections. (That's my theory, anyway. Feel free to produce data that shoots it down.)

## tl;dr

Tracking progress probably won't be that bad; not tracking progress might not be that bad. I vote to track.

# Second response to jmac

In the comments below this answer jmac points out that this proposed change would dilute the value of the badges by making them quite a bit easier to get. This is a more serious problem for the proposal, which I don't have a solution for at this time.

• Thanks for replying (+1). That would be very positive (IMO). Not sure only, if the routines to accomplish this criteria would be easy to implement. – Andre Silva Jul 31 '13 at 1:15
• @AndreSilva Votes must be identifiable at some level; the system knows to revoke the right ones when there's serial voting. Shouldn't be too hard to remove the upvote by \$OP_USER when checking for Unsung Hero. – michaelb958 Jul 31 '13 at 1:44
• Ok, perfect then. – Andre Silva Jul 31 '13 at 1:52
• Disagree for the reasons AsheeshR eloquently stated – jmac Jul 31 '13 at 2:59
• @jmac I have expanded my answer to respond; see how it looks now. – michaelb958 Jul 31 '13 at 4:31
• It looks like 34/42 of my accepted answers with upvotes were upvoted by the OP. This would change my percentage of posts for Unsung hero from it's current 31% to something like 87%. – jmac Jul 31 '13 at 4:57
• @jmac Nice destroying my "difficult to determine by other means". :) More seriously, though, that is what the proposal is all about. How many people would go from 6% to 28%? – michaelb958 Jul 31 '13 at 4:59
• The question is where do we draw the line? This is like removing the 25%, or decreasing it from 25% to 10% or somesuch. Should this be one of those situations where moonlighting in an unpopular tag occasionally gets you the badge? – jmac Jul 31 '13 at 5:06
• @jmac I'm not seeing the mathematical/statistical basis for that. You'd still need, at minimum, 10 (11?) answers either at zero score, or at one score and with an OP upvote. – michaelb958 Jul 31 '13 at 5:15
• I spend 100% of my time in unpopular tags. 31% of my accepted answers are 0-score. If we changed this counting method, that number would leap from being barely about the Unsung Hero threshold of 25%, to being dramatically over it (85%). This is the equivalent of keeping the scoring the same (zero answers only) and dropping the % of total answers from 25% down to 10% (give or take) in the sense of how much more accessible the badge would be. – jmac Jul 31 '13 at 5:42
• The point is that making this change will make the badge dramatically easier to get in the same way lowering the % of your total accepted questions from 25% to 10% would. It would dilute the "value" of the badge by making it easier to get/game for those who felt the need, without increasing the contributions of anyone. It is just redrawing the line for the badge so more people can have it. – jmac Jul 31 '13 at 5:43
• See update. cc @jmac – asheeshr Jul 31 '13 at 7:26
• @jmac, why are you saying to shrink the percentage to 10%?. What I'm proposing (asking for opinions) is: today you have 132 answers, 30 answers accepted and with +1 score, 19 accepted answers with zero score. In the current criterion, one would not be a Unsung (14.39% of zero score accepted answers). In the new scenario the percentage would be 37% (49/132), and one would get the badge. It would increase the number of badges awarded (but not deliberately). I am stating that to consider the OP's upvote is legitimate and fair (the question does not get more attention if it has one upvote). – Andre Silva Jul 31 '13 at 13:16
• @AndreSilva Unsung Hero is counted as 25% of accepted answers of which I have 61 or so. Half of my answers are unaccepted, not because they are poor, but because many people just hit and run question. That's why there's the badge. Perhaps before suggesting, you look at the query I posted and understand the badge criteria? The 10% is saying that I would go from 31% to 85% or so with just this change. That's an absurd increase in the likelihood of getting the badge even if I posted in popular questions. – jmac Jul 31 '13 at 13:34
• – jmac Jul 31 '13 at 23:55

Unpopular tags have really bad track records with getting answers accepted, let alone accepted with upvotes. The question volume is low, there are many hit-and-run questions in general, and very little in the way of alternative resources on the net.

Having valuable information not available on other resources located on Stack Overflow greatly increases the value of SO as a resource. People looking for the answers to questions will find them on SO high up in Google searches, and that will grow the knowledge base on SO and hopefully start the ball rolling toward creating a community of experts (or at least enthusiasts).

I joined to get guidance on those unpopular tags, hoping someone would know. Instead I ended up answering my own question and providing 132 answers to questions from other people.

While I don't care much about reputation or badges, it still felt nice to have a nice shiny gold badge awarded for what felt like (and still does feel like...) pretty lonely work. I think the badge works fine as-is and gives those of us fighting our shadows a twisted satisfaction when we rack up yet another 0-score accepted answer.

Some stats. I have 61 accepted answers according to this query. Of those 61 accepted answers, 42 have upvotes, and 34 seem to have been upvoted by the original poster (they had an upvote around the same time the post was marked as answered). Still trying to figure out how to use the to create a query to get the reputations of people whose questions I have answered, but I am finding myself incapable.

If this query were implemented, I would go from being 31% unsung to being over 85% unsung. I think this would dramatically change the concept of the badge from promoting questions from users without the reputation to upvote to incentivizing answers in unpopular tags.

• "gives those of us fighting our shadows a twisted satisfaction when we rack up yet another 0-score accepted answer" well put – apaul Jul 31 '13 at 2:21
• Thank you @jmac. Congratulations for being an Unsung-Hero. Let's suppose some OPs had upvoted your answers, but the questions still would be low traffic (e.g. because the unpopular tag as you cited), Would not still be a lonely work? Regardless the OP upvote you would continue to be a true Unsung-Hero. – Andre Silva Jul 31 '13 at 2:21
• This is an accurate explanation for why the badge exists +1 – asheeshr Jul 31 '13 at 2:41
• @AndreSilva Many of the OPs never come back to StackOverflow. Here is the reputation of the last 10 users I answered (minus upvotes for the question asked): 101, 20, 1, 110, 1, 2737, 7, 138, 94, 4892. So 20% of my last 10 questions were asked by 100% new users. I would write a query to see the reputation of askers whose questions I've answered, but I haven't the foggiest idea how. – jmac Jul 31 '13 at 2:42
• I like upvotes by the OP because I am not a professional programmer. I dabble. An upvote tells me it worked/is correct, which is nice. If you look at my answers, you will notice that I usually test my code (provide a working sample), show the output in the post, and put real effort in to making sure it solves the issue. Seeing that someone appreciates that (with an accepted answer or upvote) is incredibly nice. As is, 44 out of my 132 answers are 0 score, no accepted answer. After putting in effort, ANY attention is nice and makes it less lonely. – jmac Jul 31 '13 at 2:48
• Ok, not 100% of new users, right?. The point here is that considering the upvote that comes from the OP toward the badge brings no harm to the current status (IMO). However, I think the argument we are discussing would strengthen the attitude the badge tries to encourage (less gamification and more consensus by community). But the way this is going, it has been as much controversial as the other Unsung-Hero threads. – Andre Silva Jul 31 '13 at 2:50
• @jmac, no objection (agree). – Andre Silva Jul 31 '13 at 2:53
• @AndreSilva I'm saying that it isn't 100% new users who can't vote, and yet I still earned the badge without any hassle whatsoever. So it seems like your post is more, "I want to expand the number of people to get the badge" rather than realizing that those of us who earned the badge did so with little effort because of the nature of actually contributing to unpopular tags. I don't care if the OP's vote is included or excluded, I care more that it sounds like an attempt to make it easier to game a badge. – jmac Jul 31 '13 at 2:54
• Yes, I think more people would earn the Unsung, but not so much more. I am just saying that if you answer low traffic questions, you can have your answer accepted and upvoted by the OP and no one else. And one would still merit the badge. That's all. – Andre Silva Jul 31 '13 at 3:08
• @AndreSilva But the question becomes, "Why do we want more people to earn the badge?" If this will somehow give new users/unpopular tags more exposure, great! But if it just gives current users badges for doing absolutely nothing different from what they're doing now, it is just an elaborate attempt to game a badge that's tough to get as-is (because it requires rather thankless work long-term). – jmac Jul 31 '13 at 6:26
• @jmac, the question is about improving the objective for what the badge was created for (one might disagree of the proposal, though). The arguments are written in question body. I am saying why can't the answer be considered "unsung" if the OP upvotes it (I am saying there is a very soft difference between zero accepted answer and +1 accepted answer (again, one might disagree). – Andre Silva Jul 31 '13 at 12:47

I would guess that the badge is set up to encourage people to take the time to help new users who don't yet have the reputation to up-vote.

Why do this, you ask? Well, one of the much-belabored issues of the SO model is that rep is more easily accumulated by facile participation in shallow, accessible, popular topics, while successful handling of difficult, high-expertise-required issues often enough goes unrewarded. This badge attempts to act against that tendency, interpreting someone who writes a lot of accepted-but-not-voted-on answers as actively helping people solve "unpopular" problems — likely, in large part, problems from new, low-rep users who lack even the power to upvote a good answer to their question, and so may go ignored by more rep-oriented users.

I think we should leave it be. New users need answers more than we need easy badge opportunities.

If the answer was upvoted and accepted you received the 15 rep from the accept and another 10 rep from the upvote, that's a grand total of 25 rep, your efforts have been rewarded, move on.

The spirit of UNSUNG Hero, notice the emphasis on unsung, is to offer some recognition to those who labor without the extra incentive.

• Thanks for the answer. The requirement for upvoting is just 15 points of reputation. It is a very small window. Usually if the user is seeking to engage in the community he/she will be able to upvote very soon. Users who do not reach 15 points, generally asks and go away (without accepting). – Andre Silva Jul 31 '13 at 1:42
• @AndreSilva that's exactly my point... If people look at brand new users as if they're not likely to upvote or accept whats the "motivation" to answer their questions? That's where the Unsung Hero badge provides "motivation" – apaul Jul 31 '13 at 1:49
• Ok, but how one would feel motivated if they already expected to not have the answer accepted?. I can be wrong, but for what I have seen the zero score accepted answers do not come from <15 rep users. And, even if they come why would be a bad thing to consider the upvote that comes from the OP? – Andre Silva Jul 31 '13 at 1:56
• @AndreSilva If the answer was upvoted and accepted you received the 15 rep from the accept and another 10 rep from the upvote, that's a grand total of 25 rep, your efforts have been rewarded, move on. The spirit of UNSUNG hero, notice the emphasis on unsung, is to offer some recognition to those who labor without the extra incentive. – apaul Jul 31 '13 at 2:03

Jmac's answer provides an accurate explanation for why the Unsung Hero batch exists in the way that it does.

One major aspect that you are missing is data. How do you know that the majority of askers upvote answers? Voting is anonymous and hence this cannot be accurately measured by any user (only devs can). I, for one, very rarely accept and upvote the same answer.

What is the point to not consider the +1 (upvote) that generally comes from the OP?

• A lot of people are interested in measuring their progress towards badges. Currently, this can easily be measured by observation, and also through SEDE. If asker upvotes were to be not counted then the ability to measure progress would be lost (why? see next point).

• If progress measurement were possible then it would be very easy to figure out whether the OP voted you up or not. This is a direct break in the anonymity of voting. If anonymity was lost, even for just asker upvotes, we would next have comments all over SO asking and demanding for upvotes. This will just be even more troublesome for new users.

• Users may start ignoring the questions from users whom they may know to not upvote their answers. We would have the same problems that were being caused by the display of the accept rate.

• If progress measurement is not available, then there are two side effects. One, a loss of possible motivation towards acquiring the badge as mentioned in comments. The other bigger issue is the surge in meta posts asking why they havent earned a badge, what are the criteria, i should have the badge, etc. This can in itself become a problem.

It has been mentioned that tracking statistics will be tough. I disagree.

• Per question stats can be easily tracked. Look at the progress percent before answering. Then have a look after the OP responds to your answer. If no change, that means the OP hasn't voted and will probably not in the near future.

• As far as aggregate statistics are concerned, if you happen to be answering questions in a tag where you are not expecting/getting votes from people other than the OP or more than 1 vote on answers, then there is a very high probability that very few users participate in that tag. What this means is that you will in all likelihood start recognizing users pretty quickly, and so will others. Communities tend to form around such tags frequently and users not voting will get identified pretty quickly through comment threads.

• Thanks for the answer. First part, I did not say the majority of askers upvote the question ( I said it is random). What I said is that, when there is just one upvote on the answer, the most cases this upvote comes from the OP. – Andre Silva Jul 31 '13 at 2:58
• @AndreSilva I would tend to disagree with that. A lot of my accepted 0-vote questions get an upvote months after I made it (months after the answer was accepted). I could go through my reputation history to confirm if you'd like, what the ratio is of accepted answer+upvote at the same time, vs. at different times. – jmac Jul 31 '13 at 3:00
• Not necessary, I have no further comments on your answer. And you are right, I do not have data (low rep). I am just guessing, and I can be wrong. – Andre Silva Jul 31 '13 at 3:02
• I disagree with the premise of your argument: losing the ability to track badge progress is not reason enough to discard this suggestion. You're correct that progress tracking would have to be disabled, but so what? That seems like a perfectly reasonable tradeoff to make the badge actually attainable and add consistency to the kinds of user behavior encouraged. – Esoteric Screen Name Jul 31 '13 at 3:08
• @EsotericScreenName The loss of the ability to track progress would dissuade people from pursuing it, which would tend to go against the very reason the badge was created in the first place, no? – jmac Jul 31 '13 at 3:56
• @jmac I don't know. I just kind of got an unexpected Tenacious badge one day. – michaelb958 Jul 31 '13 at 4:01
• @michaelb958 That is a good point -- I didn't track my progress either and was more blindsided by the badge, but I don't think everyone necessarily shares my approach (or yours) and some people would want the tracking... – jmac Jul 31 '13 at 4:09
• Please, attempt the question refers to "maximum" of +1 score. If the answer has zero score, but it is accepted it is ok too. Accepting the question continues to be the main issue. It happens the OP will not continue to be asked "please, accept my answer but do not upvote it". – Andre Silva Jul 31 '13 at 14:06

Vote up is a privilege requiring 15 rep. If the asker of the question has less, they simply can't upvote.

In low traffic tags (where upvotes are infrequent), askers may have trouble getting there. One may ask two good questions (those who put effort into "doing their homework" prior to asking, tend to have less questions), get at most two upvotes from answerer, totaling at +10, and still be not there.

• Per my recollection, this was the case eg in tag in about 2009-2010, even reasonably well written questions weren't getting much upvotes, simply because nobody else was interested besides asker.

From this perspective Unsung Hero motivates answering questions from low rep users who can't yet upvote (they can accept though, and this is supposed to somehow indicate answer quality).

It is also worth noting that in low traffic tags, newcomers may also lack experience / examples of how to use Stack Overflow, so even after acquiring required reputation, they might still not know when, how and why to upvote.

• IMO, the same way new users can lack experience to upvote, they can lack to accept an answer. I agree unsung motivates people answering questions from new users, but it also motivates answering on threads with low traffic/attention from all community. Thanks for replying. – Andre Silva Jul 31 '13 at 13:22


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# plaintext decrypted form a cipher text using key = 3 in Caesar Cipher and that plain text with the following ciphertext to get the key [closed]

how to use the plaintext from that was decrypted form a cipher text using key = 3 in Caesar Cipher and using that plain text with the following ciphertext to get the key.

• for the second exercise: k=4, you have to arrange ciphertext in rail fence and then u can get plain text. – SSA Jan 17 at 11:20

• Take original plain text as " PLEASEEECUSEMYDEARAUNTSALLY", note that I've ignored the 'X', so that rail fence with key(row depth) = 4 cipher text will be "PEMALLEEEYRUALESCSDANSYAUET" now using Caesar cipher with k=4 on this new text.
• $${C= (P+k)mod26}$$ , you will get cipher as $${"TIQEPPIBICVYEPIWGWHERWCEYIX"}$$
• Here we take row wise so P --> T, E-->I, M-->Q, A-->E, L-->P,
• in second row L-->P,E-->I and so on...
• @user86282, Did you find the answer, if yes , then accept the answer. – SSA Jan 19 at 10:06

format: plain


text:
# What is the difference between beta regression and quasi glm with variance = $\mu(1-\mu)$?

First let me give some background; I will summarize my questions at the end.

The Beta distribution, parameterized by its mean $\mu$ and $\phi$, has $\operatorname{Var}(Y) = \operatorname{V}(\mu)/(\phi+1)$, where $\operatorname{V}(\mu) = \mu(1-\mu)$ is the variance function.

In a beta regression (e.g., using the betareg package in R), the regression assumes beta-distributed errors and estimates the fixed effects and the value of $\phi$.

In glm regression, it is possible to define a "quasi" distribution with a variance function of $\mu(1-\mu)$. So here the model assumes errors with the same variance function as Beta. The regression then estimates the fixed effects and the "dispersion" of the quasi distribution.

I may be missing something important, but it would seem that these two methods are essentially identical, perhaps differing only in their estimation method.

I tried both methods in R, regressing on a DV called "Similarity", which is in the interval $(0,1)$:

Call:
betareg(formula = Similarity ~ N + NK + Step_ent, data = TapData, link = "logit")

Coefficients (mean model with logit link):
Estimate Std. Error z value Pr(>|z|)
(Intercept)  0.715175   0.067805  10.547   <2e-16 ***
N           -0.063806   0.003858 -16.537   <2e-16 ***
NK          -0.362716   0.015008 -24.168   <2e-16 ***
Step_ent    -0.696895   0.070233  -9.923   <2e-16 ***

Phi coefficients (precision model with identity link):
Estimate Std. Error z value Pr(>|z|)
(phi)  10.6201     0.2084   50.96   <2e-16 ***
---
Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

Type of estimator: ML (maximum likelihood)
Log-likelihood:  3817 on 5 Df
Pseudo R-squared: 0.2633
Number of iterations: 18 (BFGS) + 1 (Fisher scoring)

Call:
glm(formula = Similarity ~ N + NK + Step_ent, family = quasi(link = "logit",
variance = "mu(1-mu)"), data = TapData)

Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept)  0.777451   0.069809  11.137   <2e-16 ***
N           -0.069348   0.003983 -17.411   <2e-16 ***
NK          -0.364702   0.016232 -22.468   <2e-16 ***
Step_ent    -0.704680   0.072491  -9.721   <2e-16 ***
---
Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

(Dispersion parameter for quasi family taken to be 0.0838547)

Null deviance: 566.25  on 4974  degrees of freedom
Residual deviance: 422.76  on 4971  degrees of freedom
AIC: NA

Number of Fisher Scoring iterations: 4

The coefficients of the two models are similar, as are their standard errors. The $\phi$ parameter is also similar: I assume that the Dispersion parameter (as reported by glm) and $\phi$ have the following relationship $\phi = 1/\text{Dispersion} - 1$, in which case they are 10.6201 and 10.9254, respectively.

However, none of these values is identical.

Is this because the only thing that actually differs in the two methods is their estimation procedure? Or is there some more fundamental difference I am missing? Also, is there any reason to prefer one method over the other?

• It sounds like you rediscovered fractional logistic regression... Sep 23, 2017 at 17:32

You're correct that the mean and variance functions are of the same form.

This suggests that in very large samples, as long as you don't have observations really close to 1 or 0 they should tend to give quite similar answers because in that situation observations will have similar relative weights.

But in smaller samples where some of the continuous proportions approach the bounds, the differences can grow larger because the relative weights given by the two approaches will differ; if the points that get different weights are also relatively influential (more extreme in x-space), the differences may in some cases become substantial.

In beta-regression you'd be estimating via ML, and in the case of a quasibinomial model - at least one estimated in R, note this comment in the help:

The quasibinomial and quasipoisson families differ from the binomial and poisson families only in that the dispersion parameter is not fixed at one, so they can model over-dispersion. For the binomial case see McCullagh and Nelder (1989, pp. 124–8). Although they show that there is (under some restrictions) a model with variance proportional to mean as in the quasi-binomial model, note that glm does not compute maximum-likelihood estimates in that model. The behaviour of S is closer to the quasi- variants.

I think in betareg you can get $h_{ii}$ values, and you can as well for GLMs, so at the two fitted models you can compare an approximation of each observation's relative influence (/"weight") on its own fitted value (since the other components of the ratio of influences should cancel, or nearly so). This should give a quick sense of which observations are looked at most differently by the two approaches. [One might do it more exactly by actually tweaking the observations one by one and seeing the change in fit per unit change in value]

Note that the betareg vignette gives some discussion of the connection between these models at the end of section 2.

• I assume by "samples" you are referring to observations of successes and failures? My DV "Similarity" is not a proportion of successes; it is the cosine similarity or two very high-dimensional vectors, and is bounded between 0 and 1 with no possibility of a value of 0 or 1. The beta distribution seems a reasonable choice for such data. The reason I am interested in transferring over to glm is because I wish to add a random effect too, which is not possible in betareg. I am trying to ascertain whether, in a scenario such as mine, quasi with mu(1-mu) is essentially the same. Aug 22, 2016 at 4:08
• I mean sample in the ordinary statistical sense, but I definitely don't mean a sample of counts. I mean a sample of continuous values between 0 and 1 (which are generally proportions for beta regression). If I was talking about modelling count proportions I wouldn't be talking about issues with being "close to" 0 or 1, I'd be talking about the larger problems with being exactly at 0 and 1. Aug 22, 2016 at 4:23
• Thanks for the clarification. I am probably missing something obvious, but it's not clear to me why sample size (in the traditional sense you mean) is relevant here, or why values close to 0 or 1 would be treated differently in the beta versus pseudo glm approach. Can you expound a little? Aug 22, 2016 at 4:55
• @Andrew With increasing sample sizes, log-likelihoods become more nearly quadratic near the maximum/estimates for these models become closer to multivariate Gaussian, and in that case, moments to second order tend to determine behavior. That's why with large $n$ they should tend to give the same answers. However, the functions being optimized do actually differ at finite sample size - with relative differences becoming more important at the extremes. As a result, if you get observations really close to 0 or 1, it may make a difference to the estimates, particularly if the $n$ isn't so large. Aug 22, 2016 at 5:50


text:
• Corpus ID: 251765124

# Algebraic independence of topological Pontryagin classes

@inproceedings{Galatius2022AlgebraicIO,
title={Algebraic independence of topological Pontryagin classes},
author={S{\o}ren Galatius and Oscar Randal-Williams},
year={2022}
}
• Published 24 August 2022
• Mathematics
. We show that the topological Pontryagin classes are algebraically independent in the rationalised cohomology of B Top( d ) for all d ≥ 4.

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We prove that in dimensions not equal to 4, 5, or 7, the homology and homotopy groups of the classifying space of the topological group of diffeomorphisms of a disk fixing the boundary are finitely

### The computation of even dimension surgery groups of odd torsion groups

(1978). The computation of even dimension surgery groups of odd torsion groups. Communications in Algebra: Vol. 6, No. 14, pp. 1393-1458.

### On diffeomorphisms of even-dimensional discs

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We determine $\pi_*(BDiff_\partial(D^{2n})) \otimes \mathbb{Q}$ for $2n \geq 6$ completely in degrees $* \leq 4n-10$, far beyond the pseudoisotopy stable range. Furthermore, above these degrees we

### Diffeomorphisms of discs and the second Weiss derivative of BTop(-)

• Mathematics
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We compute the rational homotopy groups in degrees up to approximately $\tfrac{3}{2}$d of the group of diffeomorphisms of a closed d-dimensional disc fixing the boundary. Based on this we determine

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One of the great achievements of contemporary mathematics is the new understanding of four dimensions. Michael Freedman and Frank Quinn have been the principals in the geometric and topological

### Milnor, Groups of homotopy spheres

• I, Ann. of Math. (2)
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### Surgery on compact manifolds, second ed

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• of Math
• 1963


text:
# $L \in RE/R$ such that $L^R \cup L \in R$

Prove/disprove: $\exists L \in RE/R$ such that $L^R \cup L \in R$

Where in my context, $R$ is the turing decidable, and $RE$ is the recursively enurmable. I tried to find such an $L$ but couldn't. What I know for sure is that I need a language in $RE/R$ such that $L \cup L^R = \Sigma^*$, or am I also wrong?

• Is $L^R$ the set of all reversed words of $L$, or is this something else? – G. Bach Jan 24 '14 at 23:35
• Yes. Reversed words of $L$ – TheNotMe Jan 24 '14 at 23:46
• Do you use $/$ to denote set difference or right quotient? – Raphael Jan 25 '14 at 16:25

First: Such an $L$ exists and $L \cup L^R$ is not necessarily $\Sigma^*$.
Try $L \cup L^R = \{a^nb^n\} \cup \{b^na^n\}$.
For each $n$, $L$ should contain exactly one of the words $a^nb^n$, $b^na^n$. How can you do this so that $L$ is in $RE/R$?
Let $K$ be any set in $RE/R$ and let $L = \{a^nb^n \mid n \in K\} \cup \{b^na^n \mid n \notin K\}$.


text:
# Magnetic Effect

1. MAGNETIC EFFECT OF CURRENT
Hans Oersted, in 1820, first discovered that when an electric current is passed through a conducting wire, a magnetic field is produced around it. If a compass needle is kept in the vicinity of the current carrying wire, the needle is found to deflect in a definite direction. If the direction of current in the wire is reversed, Then the direction of deflection of the needle is reversed.
AB is a wire lying in the north-south direction and connected to a battery through a rheostat and a tapping key. A compass needle is kept just below the wire. When the key is open i.e. no current is passed through the wire, the needle shows no deflection and points in the N-S direction (i.e. remains parallel to the wire)as shown in figure(a).

When the key is pressed and current passes in the wire in the direction A to B (i.e. from south to north and the north pole (N) of the needle deflects towards the west as figure (b). Thus a current (or moving charge) produces a magnetic field. When the direction of current in the wire is reversed by reversing the terminals of the batter, the north pole of the needle deflects towards the east as figure(c).

NOTE : If the compass needle is kept just above the wire, the deflection will be as shown in figure (d) and (e) for the direction (e) for the direction of current from A to B and from B to A respectively.
(a) Magnetic Field due to a Straight Current Carrying Wire
When a current is passed through a conducting wire, a magnetic field is produced around it. The direction of magnetic field due to a straight current carrying wire can be mapped by means of a small compass needle or by iron fillings.
Take a sheet of smooth cardboard with a hole at the centre. Place it horizontally and pass a wire vertically through the hole, Sprinkle some iron fillings on the cardboard and pass an electric current through the wire. Gently tap the cardboard. We find that the iron filling arrange themselves in concentric circles around the wire as shown in figure.
If a small compass needle is kept anywhere on the board near the wire, the direction in which the north pole of the needle points gives the direction of the magnetic the magnetic field (i.e., magnetic lines of force) at that point.

The magnetic lines of force form concentric circles near the wire, with their plane perpendicular to the straight conductor and with their centers lying on its axis. if the direction of current in the wire is reversed, the direction of lines of force is also reversed.
On increasing the strength of current in the wire, the lines of force becomes denser and iron fillings are arranged in circles upto a larger distance from the wire, showing that the magnetic field strength has increased.
(i) Magnitude of magnetic field produced by a straight current-carrying conductor : The magnitude of magnetic field (or strength of magnetic field ) B produced by an infinitely long conductor in vacuum at a distance r from it, it given by :-
B = $\frac{{\mathrm{\mu }}_{0}\mathrm{I}}{2\mathrm{\pi r}}$ B = Magnetic field strength      ${\mu }_{0}$ = Permeability of vacuum (a constant)
I = Current (flowing in conductor) and
r = Distance from the conductor (where magnetic field is measured).
The unit of magnetic field B is tesla which is denoted by the symbol T (1 tesla is equal to 1 Newton per ampere per meter). Permeability of vacuum ${\mu }_{0}$ is $4\mathrm{\pi }×{10}^{–7}$ tesla metre per ampere.
(ii) Direction of magnetic field :
The direction of magnetic field (lines of force) produced due to flow of current can be known by the following rules :
(A) Maxwell’s cork screw rule
Imagine a right handed cork screw lying with its axis coincides with the current carrying wire. It is now rotated such that it advances in the direction of the current, the direction in which the screw rotates gives the direction for the magnetic lines of force.

(B) Right hand thumb rule
If we hold the current carrying conductor in the right hand such that the thumb points in the direction of current, the fingers encircle the wire in the direction of magnetic lines of force.

(C) Ampere’s swimming rule
Imagine a man swimming along the wire in the direction of current (such that the current enters at his feet and leaves him at his head) facing towards a magnetic needle kept underneath the wire, then the magnetic field produced in such that the north pole of the needle will be deflected towards his left hand.

(b) Magnetic Field due to Circular Coil Carrying Current
A piece of wire bent in the form of a ring (or coil) is passed through a horizontal cardboard C at two points P and Q at the opposite ends of a diameter of the ring and then some iron fillings are scattered on the cardboard. The ends of the coil are connected to a battery through a rheostat and a key.
When a strong electric current is passed through the coil by closing the key and the cardboard is gently tapped we find that the iron filing arrange themselves in a definite pattern representing the magnetic lines of force due to the current carrying coil.

Direction of magnetic field is found by applying the right hand thumb rule to each section of the coil and we find that the concentric lines of force pass through the coil in the same direction. Further more that :
(i)  The magnetic lines of force are nearly circular near the wire.
(ii) Within the space enclosed by the wire, the lines of force are in the same direction.
(iii) Near the center of the coil, the lines of force are nearly parallel and the magnetic field may be assume to be practically uniform for a small space around the centre.
(iv)
At the centre, the lines of force are along its axis and at right angle to the plane of the coil.
(v)
The magnetic field strength is increased if the number of turns in the coil is increased or the strength of current in the coil is increased.
Since the magnetic lines of force through the coil point in the same direction, hence one face of the coil acts as a large area of north polarity because it is sending out magnetic lines of force and the other face acts as a large area of south polarity as magnetic lines of force are entering it. thus, the coil has a magnetic field similar to a magnetised iron disc of same radius as that of the coil.
The polarity of the faces of the coil depends on the direction of current and is determined by the clock rule. Looking at the face of the coil, if the current around the face is in an anticlockwise direction, the face has north polarity, while if the current at that face is in the clockwise direction, the face has south polarity. This can e tested by using a compass needle.

The magnitude of magnetic field B produced by a current-carrying circular wire at its centre is :
(i) directly proportional to the current I passing through the circular wire and
(ii) inversely proportional to the radius r of the circular wire.
i.e.  B $\propto$ I         and B $\propto \frac{1}{\mathrm{r}}$
Magnetic field,  B = $\frac{{\mathrm{\mu }}_{0}\mathrm{I}}{2\mathrm{r}}$
Formula which we have given above is applicable when there is only one turn of a circular wire. If we have circular coil having N turns of wire, then the magnetic field will become N times. Thus, the magnetic field at the centre of a circular coil of N turns having radius r and carrying current I is given by
$\mathrm{B}=\frac{\mathrm{N}×{\mathrm{\mu }}_{0}×\mathrm{I}}{2\mathrm{r}}$
Magnetic field produced by a circular coil carrying current is directly proportional to both, number of turn (N) and current (I), but inversely proportional to its radius (r). Thus, the strength of magnetic field produced by a current carrying circular coil can be increased by (i) increasing the number of turns of wire in the coil, (ii) increasing the current following through the coil and (iii) decreasing the radius of the coil.
(c) Magnetic Field due to a Solenoid Carrying Current
If a conducting wire is wounded in the form of a cylindrical coil whose diameter is less in comparison to the length, then this coil is called a solenoid (it looks like a helical spring).
The magnetic field lines in a solenoid, through which current is passed, are as shown in figure.

The magnetic field, thus produced, is very much similar to that of a bar magnet and one end of the coil acts like a magnetic north pole while the other acts like a south pole.
The lines of force inside the solenoid are nearly straight and parallel to the axis of the solenoid.
A strong magnetic field can be obtained by increasing the current strength.
The magnetic field is increased if the number of turns in the solenoid of given length is increased.
The magnetic field is also increased if soft iron core is kept along the axis of the solenoid.
Thus a current carrying solenoid behaves like a bar magnet with fixed polarities at its ends.

The strength of magnetic field produced by a current carrying solenoid depends upon :
(i) The number of turns in the solenoid : Larger the number of turns in the solenoid, greater will be the magnetic field produced.
(ii) The strength of current in the solenoid : Larger the current passed through solenoid, stronger will be the magnetic field produced.
(iii) The nature of “core material” used in making solenoid : The use of soft iron rod as core in a solenoid produced the strongest magnet.
Magnetic field inside the solenoid is :
B = ${\mathrm{\mu }}_{0}\mathrm{nI}$ n I        [Here in is number of turns per unit length]
At the ends of the solenoid the magnetic field :
${\mathrm{B}}_{\mathrm{end}}=\frac{1}{2}{\mathrm{\mu }}_{0}\mathrm{nI}$

2.  PERMANENT AND TEMPORARY MAGNETS
The degree to which magnetism is retained by a given piece of iron depends entirely upon its constitution. Steel retains the largest amount while soft iron retains the least. Therefore pieces of steel are employed to prepare permanent magnets, whereas soft iron is used for preparing temporary magnets, i.e., magnets that retain their magnetism only as long as the current flows in the magnetising coil. They lose their magnetism as soon as the current is switched off. Such magnets are known as electromagnets.
(a) Electromagnet
An electric current can be used for making temporary magnets known as electromagnets. As electromagnet works on the magnetic effect of current. When current is passed through a long coil called solenoid, a magnetic field is produced. It has been found that if a soft iron rod called core, is placed inside a solenoid then the strength of magnetic field becomes very large because the iron core gets magnetised by induction. This combination of a solenoid and a soft iron core is called an electromagnet.
Electromagnets can be made in different shapes and sizes depending on the purpose for which they are to be used.
Factors affecting the strength of an electromagnet are :
(i) The number of turns in the coil: If we increase the number of turns in the coil, the strength of electromagnet increases.
(ii) The current flowing in the coil : If the current in the coil is increased, the strength of electromagnet increases.
(iii) The  length of air between its poles : if we reduce the length of air gap between the poles of an electromagnet, then its strength increases.
For example, the air gap between the poles of straight bar type electromagnet is quite large, so a bar type electromagnet is not very strong. One the motherland the air gap between the poles of a U-shaped electromagnet is small, so it is a very strong electromagnet.
Electromagnets are used in electric bells, telegraphs, telephones and several other instruments. Since the magnetisation depends on the current flowing through the coil, it is possible to obtain very powerful electromagnets by increasing the current.
Soft iron can be easily magnetised every by a weak magnetic field, whereas steel can be magnetised only by strong magnetic field.
Less energy is required for magnetising soft iron. Soft iron loses its magnetism immediately, whereas steel retains it magnetism.
(b) Difference between a Bar Magnet (or Permanent Magnet) and an Electromagnet

 S.No. Bar magnet (or permanent magnet) Electromagnet (1) The bar magnet is a permanent magnet. An electromagnet is a temporary magnet. Its magnetism is only for the duration for which current passes through it, so the magnetism of an electromagnet can be switched on or switched off as desired. (2) A permanent magnet produces a comparatively weak force of attraction. An electromagnet can produce very strong magnetic force. (3) The strength of a permanent magnet cannot be changed. The strength of an electromagnet can be changed by changing the number of turns in its coil or by changing the current passing through it. (4) The (north-south) polarity of permanent of manget is fixes and cannot be changed. The polatiry of an electromagnet can be changed by changing the direction of current in its coil.

Permanent magnets are usually made of alloys such as carbon-steel, chromium-steel, cobalt-steel, tungsten-steel, nipermag and alonico. Nipermag is an alloy of iron, nickel, aluminum and titanium whereas ALNICO is an alloy of aluminum, nickel and cobalt. Permanent magnets of these alloyws are much more stronger than those made of ordinary steel, such strong permanent magnets are used in microphones, loudspeakers, electric clocks, ammeters, voltmeters, speedometers and many other devices.

(c) Methods of Demagnetising a Permanent Magnet
(i) Magnet can be demagnetised by :
(A) Self – demagnetisation, if the magnet is strode without using magnetic keepers.
(B) Dropping it from a height or by rough handling.
(C) Heating or hammering the magnet.
(ii) Magnet can be demagnetised by placing it within a solenoid and passing high frequency AC through it.

3.  USED OF MAGNETISM IN MEDICINE
An electric current always produces a magnetic field. Even weak ion currents that travel along the nerve cells in our body produce magnetic fields.
When we touch something, our nerves carry an electric impulse to the muscles we need to use. The impulse produces a temporary magnetic field. These field are very weak and are one billionth of the earth’s magnetic field. Heart and brain are the two main organs in the human body where the magnetic field produced is significant. The magnetic field inside the body forms the base of obtaining the images of different body parts. This is done by using a technique called Magnetic Resonance Imaging (MRI). Analysis of these images helps in medical diagnosis. Magnetism has thus, got important uses in medicine.
4. MAGINETIC FORCE
(a) Force on a Current-Carrying Conductor in a Magnetic Field
Immediately after Oersted’s discovery of electric currents producing magnetic fields and exerting forces on magnets, Ampere suggested that magnet must also exert equal and opposite force on a current-carrying conductor. When a current carrying conductor is kept in a magnetic field (not parallel to it), a force acts on it. This force is created due to the interaction of magnetic field of the current in the conductor and the external magnetic field on the conductor. As a result of this superposition, the resultant magnetic field on one side of conductor is weaker than on the other side. hence the conductor experience a resultant force in one direction.
Take a small aluminum rod AB. Suspend it horizontally by means of two connecting wires from a stand. Now, place a strong horseshoe magnet in such a way that the rod is between the two poles with the field directed upwards. If a current is now passed in the road from B to A, we will observe that the rod gets displaced. This displacement is caused by the force acting on the current-carrying rod. The magnet exerts a force on the rod directed towards the right, with the result the rod will get deflected to the right. If we reverse the current or interchange the poles of the magnet, the deflection of the rod will reverse, indicating thereby that the direction of the force acting on it gets reversed. This shows that there is a relationship among the directions of the current, the field and the motion of the conductor.

(b) Direction of Force on Current Carrying Conductor
The direction of force obtained by the Fleming’s left hand rule.
Fleming left hand rule :
Stretch the forefinger, middle finger and the thumb of you lef hand mutually perpendicular to each other as shown in figure. It the forefinger indicates the direction of the magnetic field and the middle finger indicates the direction of current, then the thumb will indicated the direction of motion (i.e., force) on the conductor.

(c) Magnitude of Force
Experimentally it is found that the magnitude of the force acting on a current carrying conductor kept in a magnetic field in direction perpendicular to it, depends on the following factors :
(i) The force F is directly proportional to the current flowing in the conductor, i.e. F $\propto$ I.
(ii) The force F is directly proportional to the intensity of magnetic field, i.e. F $\propto$ B.
(iii) The force F is directly proportional to the  length of the conductor (inside the magnetic field), i.e. F ∝ $l$ Combining these we get, F ∝ I B $l$ Or         F = K I B $l$
Where K is constant whose value depends on the choice of units. In S.I. units K = 1 and the unit of magnetic field is tesla (T). 1 tesla is equal to 1 Newton ampere-1 metre-1 or 1 Weber metre-2.
Force is directly proportional to sin $\theta$ where  is the angle between current and the direction of magnetic field. i.e. F ∝ sin $\theta$
Combining all we have F = BI$l$  sin$\theta$      or    $\stackrel{\to }{\mathrm{F}}=\mathrm{I}\left(\stackrel{\to }{l}×\stackrel{\to }{\mathrm{B}}\right)$
Special cases :
(i) When $\theta =0°$   or  $180°$,   $\mathrm{sin}\theta =0$
Force on a current – carrying conductor placed parallel or ant parallel to field is zero.
(ii)
If $\theta =90°$, sin 900 = 1, F = BI$l$  is the maximum force. Force experienced by the conductor is maximum when placed perpendicular to magnetic field.
(iii)
I of B = 0, F = 0 i.e. the coil placed in field free area doesn’t experience any force.
A moving charge in a magnetic field (direction of motion not parallel to the field direction) experiences a force called Lorentz force. Since current is due to flow of charge, therefore a conductor carrying current will experience a force.
The force acting on a current – carrying conductor placed in a magnetic field is :
F = BI$l$
Now, if a charge Q flows in time t then the current I=$\frac{\mathrm{Q}}{\mathrm{t}}$. So, writing $\frac{\mathrm{Q}}{\mathrm{t}}$ in place of I in the above equation,
we get : $\mathrm{F}=\frac{\mathrm{B}×\mathrm{Q}×l}{\mathrm{t}}$
Suppose the particle carrying the charge Q travels a  length $l$ in time t. Then the velocity v of the charged particle will be equal to $\frac{l}{\mathrm{t}}$ . Writing v in place of $\frac{l}{\mathrm{t}}$ in the above equation, we get :
Force on moving charge, F = B × q × v
Where B = Magnitude of magnetic field, Q = Charge on the moving particle and v = Velocity of the charged
particle (in metre per second). In vector notation $\stackrel{⇀}{\mathrm{F}}=\mathrm{Q}\left(\stackrel{⇀}{\mathrm{v}}×\stackrel{⇀}{\mathrm{B}}\right)$

5. ELECTROMAGNETIC INDUCTION
When an electric current is passed through a conductor, a magnetic field is produced around the conductor. Faraday thought that as a magnetic field is produced by electric current, it should be possible to produce an electric current by the magnetic field. According to him, whenever there is a change in the magnetic lines of force associated with a conductor, an electromotive force (e.m.f.) is set up at the ends of the conductor which lasts as long as the change is taking place. This phenomenon is called electromagnetic induction.
Wind an insulated copper wire on a wooden cylinder so as to form a solenoid coil. Connect the two ends of the coil to the centre of galvanometer. A magnet is placed along the axis of the coil.
(i) When the magnet is stationary, there is no deflection in the galvanometer. The pointer reads zero as shown in figure (A).
(ii) When the north pole of the magnet is brought near the coil, the current flows in the coil in direction shown in the figure (B) and the galvanometer shows the deflection towards the right.

(iii) If we stop the motion of the magnet, the pointer of the galvanometer comes to the zero position as shown in figure (C). Thus the current in the coil flows so long as the magnet is moving. If the magnet is taken away from the coil, the current again flows in the coil but in the direction opposite to that shown in figure (D) and therefore the pointer of the galvanometer deflects towards the left side.
(iv) If south pole of the magnet is brought towards the coil, the current in the coil flow in the direction opposite to that shown in figure (E) and so the pointer of the galvanometer deflects towards the left.
(v) Similar deflection is observed in the galvanometer if the magnet is kept stationary and the coil is moved.

From this experiment Faraday concluded that :
(i) The galvanometer shows a deflection (i.e. current flow in the coil) only when there is relative motion between the coil and the magnet.
(ii)
The direction of deflection is reversed if the direction of motion is reversed.
(iii)
The value of the current in the coil (i.e. deflection of the pointer) is increased by :
(A) The rapid motion of the magnet or the coil.
(B) the use of a strong magnet.
(C) increasing the area and number of turns in the coil.
When the magnet and coil are relatively at rest, the total number of magnetic lines of force due to the magnet passing through the coil (i.e. the magnetic flux linked with the coil) remains constant, therefore no e.m.f. is induced in the coil and the galvanometer shows no deflection.
When there is relative motion between the coil and magnet, the magnetic flux linked with coil changes. If the coil is moved towards the magnet, the magnetic flux through the coil increases as shown in fig. Due to change in magnetic flux linked with the coil, an e.m.f. is induced in the coil. This e.m.f. causes a current to flow if the circuit of the coil is closed.

(b) Faraday’s Laws of Electromagnetic Induction
Faraday formulated the following two laws of electromagnetic induction :
(i) Whenever there is a change in magnetic flux linked with a conductor, an em.f. is induced. The induced e.m.f. lasts so long as there is a change in magnetic flux cut by the conductor.
(ii) The magnitude of the e.m.f. induced is directly proportional to the rate of change of magnetic flux cut by the conductor. If the rate of change of magnetic flux remains uniform, a steady e.m.f. is induced. If the circuit of conductor is closed, a current flows in the conductor due to the e.m.f. induced across its ends.
(c) Direction of Induced e.m.f.

The direction of induced e.m.f. (and hence the direction of induced current) can be obtained by any of the following rules :
(i) Fleming’s right hand rule
(ii) Lenz’s law
(i) Fleming’s right hand rule :
Stretch the thumb, middle finger and the forefinger of your right hand mutually perpendicular to each other as shown in figure. If the forefinger indicated the direction of the magnetic field and the thumb indicated the direction of motion of the conductor, the middle finger will indicate the direction of induced current.

(ii) Lenz’s law : This law gives us the direction of current induced in a circuit.
According to Lenz’s law, the induced current will appear in such a direction that it opposes the change (in magnetic flux) responsible for its production.
The law refers to induced currents, which means that it applies only to closed circuits. If the circuit is open we would find the direction of induced e.m.f.
For example, in figure, when the magnet is moved towards the loop, a current is induced in the loop. The induced current produces its own magnetic field with magnetic dipole moment  oriented so as to oppose the motion of the magnet. Thus the induced current must be anticlockwise as shown in figure below.

6. GENERATOR
This is a divide which convert mechanical energy into electrical energy using the principle of electromagnetic induction. It is of two types :
AC Generator or Dynamo
When a coil (conductor) is rotated in a magnetic field, the magnetic flux linked with it changes and therefore an alternating e.m.f. is induced in the coil.
Construction : The main parts of dynamo are :-
(i) Field magnet : It is a strong horse shoe permanent magnet. An electromagnet run by a DC source can also be used for high power generators.
(ii) Armature:  It is a soft iron core on which a coil ABCD having a large number of turns of insulated copper wire is wound. This armature (or coil) is rotated rapidly in the magnetic field between the poles of the magnet.
(iii) Slip rings : The ends of the armature (or the coil) are connected to two coaxial metallic slip rings which rotate along with the coil.
(iv) Brushes :  Two brushes ${\mathrm{B}}_{1}$ and ${\mathrm{B}}_{2}$ made of carbon, press against the slip rings ${S}_{1}$ and ${\mathrm{S}}_{2}$ respectively. The external circuit (i.e. load) is connected between the other ends of brushed. The brushed ${\mathrm{B}}_{1}$ and ${\mathrm{B}}_{2}$ do not rotate along with the coil.
Working of an AC generator
Suppose that the generator coil ABCD is initially in the horizontal position. Again suppose that the coil ABCD is being rotated in the anticlockwise direction between the poles N and S of a horse-shoe type magnet.
(i) As the coil rotates in the anticlockwise direction, the side AB of the coil moves down cutting the magnetic lines of force near the N- pole of the magnet and side CD moves up, cutting the lines of force near the S-pole of the magnet. Due to this induced current is produced in the side AB and DC of the coil. On applying Fleming’s right-hand rule to the sides AB and DC of the coil, we find that the currents are in the directions B to A and D to C. Thus, the induced currents in the two sides of the coil are in the same direction and we get an effective induced current in the direction BADC.
(ii) After half revolution, the sides AB and DC of the coil will interchange their positions. The side AB will come on the right hand side and side DC will come on the left hand side. So, after half a revolution, side AB starts moving up and side DC starts moving down. As a result of this, the direction of induced current in each side of the coil is reversed after half revolution. Since the direction of induced current in the coil is reversed after half revolution so that polarity (positive and negative) of the two ends of the coil also changes after half revolution. The end of coil which was positive in the first half of rotation becomes negative in the second half. And the end which was negative in the first-half revolution becomes positive in the second half of revolution. Thus, in 1 revolution of the coil, the current changes its direction 2 times.
The alternating current (AC) produced in India has a frequency of 50 Hz. That is, the coil is rotated at the rate of 50 revolutions per second. Since in 1 revolution of coil, the current changes its direction 2 times, so in 50 revolutions of coil, the current changes its direction 2 × 50 = 100 times. Thus, the AC supply in India changes its direction 100 times in 1 second. Another way of saying this is the alternating current produced in India changes its direction every 1/100 second. That is, each terminal of the coil is positive (+) for 1/100 of a second and negative (-) for the next 1/100 of a second.

After every half revolution, each side of the generator coil starts moving in the opposite direction in the magnetic field. The side of the coil which was initially moving upwards, after half revolution, it starts moving downwards. Due to the change in the direction of motion of the two sided of the coil in the magnetic field every half revolution, the direction of current produced in them also changes after every half revolution.
(b) DC Generator (or DC Dynamo)
“DC generator” means “Direction Current generator”. That is, a DC generator produces direct current
Construction of a DC Generator :
A simple DC generator consists of a rectangular coil ABCD which can be rotated rapidly between the poles north and south of a strong horse-shoe type magnet M.
The generator coil is made of a large number of turns of insulated copper wire. The two ends of the coil are connected to the two copper half rings (or split rings) ${\mathrm{R}}_{1}$ and ${\mathrm{R}}_{2}$ commutator. There are two carbon brushes ${\mathrm{B}}_{1}$ and ${\mathrm{B}}_{2}$ which press lightly against the two half rings. When the coil is rotated, the two half rings ${\mathrm{R}}_{1}$ and ${\mathrm{R}}_{2}$ touch the two carbon brushes ${\mathrm{B}}_{2}$ and ${\mathrm{B}}_{2}$ one by one. So, the current produced in the rotating coil can be tapped out through the commulator half rings into the carbon brushes. From the carbon brusher ${\mathrm{B}}_{1}$ and ${\mathrm{B}}_{2}$. we can take the current into the various electrical appliances like radio, T.V., electric iron, bulbs, etc.

Working of a DC generator
Suppose that the generator coil ABCD is initial in the horizontal position. Again suppose that the coil ABCD is being rotated in the anticlockwise in the anticlockwise direction between the poles N and S of a horse-shoe type magnet.
(i) As the coil rotates in the anticlockwise direction, the side AB of the coil move down cutting the magnetic  lines of force near the N-pole of the magnet and side DC moves up, cutting the lines of  force near the S-pole of the magnet in figure. Due to this, induced current is produces in the sides AB and DC of the coil. On applying Fleming’s right-hand rule to the sides AB and DC of the coil we find that the currents in them are in the directions B to A and to C respectively. Thus, we get an effective induced current in the direction BADC. Due to this the brush B1 becomes a positive (+) pole and brush B2 becomes negative (-) pole of the generator.
(ii) After half revolution the sides AB and DC of the coil will interchange their positions. The side AB will come on the right hand side and start moving up whereas side DC will come on the left-hand side and start moving down. But when sides of the coil interchange their position, then the two commutator half rings ${\mathrm{R}}_{1}$ and ${\mathrm{R}}_{2}$ automatically change their contacts from one carbon brush to the other. Due to this change, the current keeps flowing in the same direction in the circuit. Te brush ${\mathrm{B}}_{1}$ will always remain positive terminal and brush ${\mathrm{B}}_{2}$ will always remain negative terminal of the generator. Thus, a DC generator supplies a current in one direction by the use of a commutator consisting of two half-rings of copper.
Difference between a DC generator an AC generator
In a DC generator we connect the two ends of the coil to a commutator consisting of two, half rings of copper. On the other hand, in an AC generator, we connect the two ends of the coil to two full rings of copper called slip rings.
7. ELECTRIC MOTOR
A motor is a device which converts electrical energy into mechanical energy. Every motor has a shaft or spindle which rotates continuously when current in passed into it. The rotation of its shafts is used to drive the various types of machines in homes and industry. Electric motor is used in electric fans, washing machines, refrigerators, mixer and grinder and many other appliances. A common electric motor works on direct current. So, it is also called DC motor, which means a “Direct Current motor”/ The electric motor which we are going to discuss now is actually a DC motor.
(a) Principle of a Motor
An electric motor utilizes the magnetic effect of current. A motor works on the principle that when a rectangular coil is placed in a magnetic field and current is passed through it, a torque acts on the coil which rotates is continuously. When the coil rotates, the shaft attached to it also rotates. In this way the electrical energy supplied to the motor is converted into the mechanical energy of rotation.
(b) Construction of a Motor
An electric motor consists of a rectangular coil ABCD of insulated copper wire, wound on a soft iron core called armature. The soft iron core has not been shown in figure to make things simple. The coil is mounted between the curved poles of a U-shaped permanent magnet is such a way that it can rotate between the poles N and S. The two ends of the coil are soldered (or welded) permanently to the two half rings X and Y of a commutator.

A commutator is a copper ring split into two parts X and Y, these two parts are insulated from one another and mounted on the shaft of the motor.
End A of the coil is welded to part X of the commutator and end D of the coil is welded to part Y of the commutator. The commutator rings are mounted on the shaft of the coil and they also rotate when the coil rotates.
The function of commutator rings is to reverse the direction of current following through the coil every time the coil just passes the vertical position during a revolution.
We cannot join the battery wire directly to the two commutator’s half rings to pass current into the coil because if we do so, then the connecting wires will get twisted when the coil rotates. So, to pass the electric current to the coil, we use two carbon strips P and Q known as brushes. The carbon brushes P and Q are fixed to the base of the motor and they press lightly against the two half rings of the commutator. The function of carbon brushes is to make contact with the rotating rings of the commutator and through them to supply current to the coil. It should be noted that any one brush touches only one ring at a time, so that when the coil rotates, the two brushes will touch both the rings one by one.
(c) Working of a Motor
Suppose that initially the coil ABCD is in the horizontal position as shown in figure. On pressing the switch, the current enters the coil through carbon brush P and commutator half ring X. The current flows in the direction ABCD and leaves via ring Y and brush Q.
(i) In side AB of the coil, the direction of current is from A to B and the direction of magnetic field is from N to S pole. So, by applying Fleming’s left hand rule to the side AB of the coil we find that it will experience a force in the upward direction.
(ii) In side DC of the coil, the direction of current is from C to D towards but the direction of magnetic field remains the same from N to S pole as shown in figure. So, by applying Fleming’s lef hand rule to the side DC of the coil, we find that. It will experience a force in the downward direction.
(iii) We find that the force acting on the side AB of the coil is in the upward direction whereas the force acting on the side DC of the coil is in the downward direction. These two equal, opposite and parallel forces acting on the two sided to the coil form a couple (torque) and rotate the coil in the anticlockwise direction.
(iv) While rotating, when the coil reaches the vertical position, then the brushes P and Q will touch the gap between the two commutator rings and current to the coil is cut off. Though the current to the coil is cut off when it is in the exact vertical position, the coil doesn’t stop rotating because it has already gained momentum due to which it goes beyond the vertical position.
(v) When the coil goes beyond the vertical position, the two commutator’s half rings automatically change contact from one brush to the other. This reverses the direction of current through the coil which, in turn, reverses the direction of forces acting on the two sides of the coil. The side AB of the coil now be one the left hand side with a downward force on it, whereas side DC of the coil will come on the right hand side with an upward force on it. In this position also a couple acts on the coil which rotates it in the same direction (anticlockwise direction). This process is repeated again and again and the coil continues to rotate as long as the current is passing. This is how an electric motors works.


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# GSEB Solutions Class 12 Statistics Part 2 Chapter 1 Probability Ex 1

Gujarat Board Statistics Class 12 GSEB Solutions Part 2 Chapter 1 Probability Ex 1 Textbook Exercise Questions and Answers.

## Gujarat Board Textbook Solutions Class 12 Statistics Part 2 Chapter 1 Probability Ex 1

Section A

Answer the following questions by selecting a correct option from the given options:

Question 1.
Which event is given by a special subset Φ of the sample space U ?
(a) Certain event
(b) Complementary event of Φ
(c) Union of events U and Φ
(d) Impossible event
(d) Impossible event

Question 2.
What is the value of P(A ∩ A’) for events A and A’ ?
(a) 1
(b) 0
(c) 0.5
(d) between 0 and 1
(b) 0

Question 3.
Which of the following options is true for any event of the sample space ?
(a) P(A) < 0 (b) 0 ≤ P(A) ≥ 1 (c) 0 ≤ P(A) ≤ 1 (d) P(A) > 1
(c) 0 ≤ P(A) ≤ 1

Question 4.
Which of the following options is not true for any two events A and B in the sample space U; where, A ⊂ B?
(a) P(A ∩ B) = P(B)
(b) P(A ∩ B) = P(A)
(c) P(A ∪ B) ≥ P(A)
(d) P(B-A) = P(B)-P(A)
(a) P(A ∩ B) = P(B)

Question 5.
What is the other name of the classical definition of probability?
(a) Mathematical definition
(b) Axiomatic definition
(c) Statistical definition
(d) Geometric definition
(a) Mathematical definition

Question 6.
Which of the following statement for probability of elementary events H and T of random experiment of tossing a balanced coin is not true ?
(a) P(T) = 0.5
(b) P(H) + P(T) = 1
(c) P(H ∩ T) = 0.5
(d) P(H) = 0.5
(c) P(H ∩ T) = 0.5

Question 7.
Which random experiment from the following random experiments has an infinite sample space ?
(a) Throwing two dice
(b) Selecting two employees from an office
(c) To measure the life of electric bulb
(d) Select a card from 52 cards
(c) To measure the life of electric bulb

Question 8.
If A ∪ A’ = U, then what type of events are A and A’ ?
(a) Independent events
(b) Complementary events
(c) Certain events
(d) Impossible events
(b) Complementary events

Question 9.
If P(A|B) = P(A) and P(B|A) = P(B), then what type of events are A and B ?
(a) Independent events
(b) Complementary events
(c) Certain events
(d) Impossible events
(a) Independent events

Question 10.
Two events A and B of a sample space are mutually exclusive. Which of the following will be equal to P(B-A)?
(a) P(A)
(b) P(B)
(c) P(A ∩ B)
(d) P(A ∪ B)
(b) P(B)

Question 11.
What is the total number of sample points in the sample space formed by throwing three six-faced balanced dice simultaneously?
(a) 62
(b) 36
(c) 6 × 3
(d) 63
(d) 63

Question 12.
If one number is randomly selected between 1 and 20, what is the probability that the number is a multiple of 5 ?
(a) $$\frac{1}{2}$$
(b) $$\frac{1}{6}$$
(c) $$\frac{1}{5}$$
(d) $$\frac{1}{3}$$
(b) $$\frac{1}{6}$$

Question 13.
If events A and B are independent, which of the following options is true?
(a) P(A ∩ B) = P(A) × P(B)
(b) P(A ∪ B) = P(A) + P(B)
(c) P(A ∪ B) = P(A) × P(B)
(d) P(A ∩ B) = P(A) + P(B)
(a) P(A ∩ B) = P(A) × P(B)

Question 14.
What is the probability of having 5 Thursdays in the month of February in a year which is not a leap year?
(a) 0
(b) $$\frac{1}{7}$$
(c) $$\frac{2}{7}$$
(d) $$\frac{3}{7}$$
(a) 0

Question 15.
If P(A) = 0.4 and P(B’) = 0.3 for two independent events A and B of a sample space, then state the value of P(A ∩ B).
(a) 0.12
(b) 0.42
(c) 0.28
(d) 0.18
(c) 0.28

Question 16.
For two events A and B of a sample space, state the event (A ∩ B) ∪ (A ∩ B’).
(a) Φ
(b) B
(c) A
(d) ∪
(c) A

Question 17.
According to the mathematical definition of probability, what is the probability of each outcome among the n outcomes of a random experiment ?
(a) 0
(b) $$\frac{1}{n}$$
(c) 1
(d) cannot say
(b) $$\frac{1}{n}$$

Section B

Answer the following questions in one sentence:

Question 1.
Give two examples of random experiment.
Two examples of random experiment are:

1. The experiment of throwing a balanced die and
2. The experiment of finding defective units from a lot of units produced.

Question 2.
Draw the Venn diagram for A – B, the difference event of A and B.
The Venn diagram for difference event A-B:

Question 3.
Define an event.
Any subset of the sample space of a random experiment is called an event. It is denoted by A, B, C ….

Question 4.
Write the sample space of a random experiment of throwing one balanced die and a balanced coin simultaneously.
The sample space of a random experiment of throwing one balanced die and a balanced coin simultaneously is obtained as follows:
U = {(1, H), (2, H), (3, H), (4, H), (5, H), (6, H), (1, T), (2, T), (3, T), (4, T), (5, T), (6, T)}
Where, H = Head; T = Tail;
1, 2, 3, 4, 5, 6 = Numbers on die.

Question 5.
Define conditional probability.
U is a finite sample space and A and B are any two events of U. The probability of event B, under the condition that event A is happened, is called conditional probability of the event B.

Question 6.
State the formula for the probability of occurrence of at least one event out of three events A, B and C.
The formula for the probability of occurrence of at least one event out of three events A, B and C is as follows:
P(A ∪ B ∪ C) = P(A) + P(B) + P(C) – P(A ∩ B) – P(A ∩ C) – P(B ∩ C) + P(A ∩ B ∩ C)

Question 7.
Define independent events.
U is a finite sample space and A and B are any two events of U. If the probability of occurrence of the event A does not depend on the occurrence or non-occurrence of the event B, then the events A and B are called independent events.

Question 8.
Write the law of multiplication of probability for two independent events A and B in a sample space.
The law of multiplication of probability for two independent events A and B in a sample space is as follows:
P(A ∩ B) = P(A) × P(B)

Question 9.
Interpret P(A|B) and P(B|A).
Interpretation of P(A|B): The conditional probability of the event A under the condition that the event B is occurred.
Interpretation of P(B|A): The conditional probability of the event B under the condition that the event A has occurred.

Question 10.
When can we say that three events A, B and C in a sample space are exhaustive?
When P(A) + P(B) + P(C) = 1, we can say that three events A, B and C in a sample space are exhaustive.

Question 11.
Arrange P(A ∪ B), P(A), P(A ∩ B), 0, P (A) + P (B) in the ascending order.
0, P(A ∩ B), P(A), P(A ∪ B), P(A) + P(B) are in the ascending order.

Question 12.
Define:
1. Random Experiment: The experiment which can be independently repeated under identical conditions and all its possible outcomes are known but it cannot be predicted with certainty which of the outcomes will appear is called a random experiment.

2. Sample Space: The set of all possible outcomes of a random experiment is called a sample space of that random experiment. It is denoted by U or S.

3. Equi-probable Events: If there is no apparent reason to believe that out of one or more events of a random experiment, any one event is more or less likely to occur than the other events, then those events are called as equi-probable events.

4. Favourable Outcomes: If some outcomes out of all the elementary outcomes in the sample space of a random experiment indicate the occurrence of a certain event A, then these outcomes are called the favourable outcomes of the event A.

5. Probability (Mathematical Definition): If out of n outcomes of the finite sample space of a random experiment which are mutually exclusive, exhaustive and equi-probable, m outcomes are favourable for an event A, then the probability of the event A, P(A) = $$\frac{m}{n}$$.

6. Probability (Statistical Definition) :
Suppose, a random experiment is repeated n times under identical conditions. If an ‘event A occurs in m trials then the relative frequency $$\frac{m}{n}$$ of the event A gives the estimate of the probability of the event A. When n tends to infinity, the limiting value of $$\frac{m}{n}$$ is called the probability of the event A. Thus,

7. Impossible Event: The special subset Φ or { } of the sample space of a random experiment is called an impossible event.

8. Certain Event: The special subset U of the sample space of a random experiment is called a certain event.

Question 13.
For two events A and B in a sample space, A ∩ B = Φ and A ∪ B = U. State the values of P(A ∩ B) and P(A ∪ B).
A ∩ B = Φ and A ∪ B = U
∴ P(A ∩ B) = 0 and P(A ∪ B) = 1

Question 14.
If two events A and B in a sample space are independent, then state the formula for P(A ∪ B).
A and B are independent events.
∴ P(A ∪ B) = P(A) + P(B) – P(A ∩ B)
= P(A) + P(B) – [P(A) ∙ P(B)]

Question 15.
If A = {x|0 < x < 1} and B = {x|$$\frac{1}{4}$$ ≤ x ≤ 3}, then find A n B.
A = {x|0 < x < 1} = {$$\frac{1}{4}, \frac{1}{2}, \frac{3}{4}$$}
B = {x|$$\frac{1}{4}$$ ≤ x ≤ 3} = {$$\frac{1}{4}, \frac{1}{2}, \frac{3}{4}$$, ……, 1, 2, 3}
∴ A ∩ B = {x|$$\frac{1}{4}$$ ≤ x < 1}

Question 16.
For two independent events A and B, P (A) = 0.5 and P(B) = 0.7. Find P(A’ ∩ B’).
A and B are independent events. So A’ and B’ are also independent events. P(A) = 0.5, P(B) = 0.7 are given.
∴ P(A’ ∩ B’) = P(A’) ∙ P(B’)
= [1 – P(A)] [1 – P(B)]
= [1 – 0.5] [1 – 0.7]
= [0.5 × 0.3]
= 0.15

Question 17.
If P(A) = 0.8 and P (A ∩ B] = 0.25, find P(A – B).
P(A) = 0.8, P(A ∩ B) = 0.25 are given.
∴ P (A – B) = P(A) – P(A ∩ B)
= 0.8 – 0.25 = 0.55

Question 18.
If P(A) = 0.3 and P(A ∩ B) = 0.03, find P(B|A).
P(A) = 0.3, P(A ∩ B) = 0.03 are given.
∴ P(B|A) = $$\frac{P(A \cap B)}{P(A)}$$ = $$\frac{0.03}{0.3}$$ = 0.1

Question 19.
If P(A) = P(B) = K for two mutually exclusive events A and B, find P(A ∪ B).
P (A) = P(B) = k.
A, B are mutually exclusive events
∴ P(A ∩ B) = 0
Now, P(A ∪ B) = P(A) + P(B) .
= k + k = 2k

Question 20.
If P(A’ ∩ B) = 0.45 and A ∩ B = Φ, find P(B).
P (A’ ∩ B) = 0.45 and A ∩ B = Φ are given.
∴ P(A ∩ B) = 0
Now, P(A’ ∩ B) = P(B) – P(A ∩ B)
∴ 0.45 = P(B) – 0
∴ P(B) = 0.45

Question 21.
Two events A and B in a sample space are mutually exclusive and exhaustive. If P(A) = $$\frac{1}{3}$$, find P(B).
A and B are mutually exclusive and exhaustive events.
∴ P(A) + P(B) = 1 (Putting P(A) = $$\frac{1}{3}$$)
$$\frac{1}{3}$$ + P(B) = 1
∴ P(B) = 1 – $$\frac{1}{3}$$ = $$\frac{2}{3}$$

Question 22.
2 % items in a lot are defective. What is the probability that an item randomly selected from this lot is non-defective?
A = Event that an item is defective
∴ P(A) = 2 % = $$\frac{2}{100}$$ = 0.02
A’ = Event that an item is non-defective
∴ P(A’) = 1 – P(A) = 1 – 0.02 = 0.98

Question 23.
State the number of sample points in the random experiment of tossing five balanced coins.
The number of sample points in the random experiment of tossing five balanced coins is n = 25 = 32.

Question 24.
State the number of sample points in the random experiment of tossing one balanced coin and two balanced dice simultaneously.
The number of sample points in the random experiment of tossing one balanced coin and two balanced dice simultaneously is n = 21 × 62 = 2 × 36 = 72.

Question 25.
Is it possible that P (A) = 0.7 and P (A ∪ B) = 0.45 for two events A and B in a sample space ?
P (A) = 0.7 and P(A ∪ B) = 0.45 is not possible because P(A ∪ B) < P(A).

Question 26.
Two cards are selected one by one with replacement from 52 cards. State the number of elements in the sample space of the random experiment.
The number of elements in the sample space of this random experiment of drawing two cards one by one with replacement from 52 cards is n = 52C1 × 52C1 = 52 × 52 = 2704

Question 27.
For two independent events A and B, P(B|A) = $$\frac{1}{2}$$ and P(A ∩ B) = $$\frac{1}{5}$$. Find P(A).
P(B|A) = $$\frac{1}{2}$$, P(A ∩ B) = $$\frac{1}{5}$$, P(A) = ?
P(B|A) = $$\frac{P(A \cap B)}{P(A)}$$
∴ $$\frac{1}{2}$$ = $$\frac{\frac{1}{5}}{\mathrm{P}(\mathrm{A})}$$
∴ $$\frac{1}{2}$$p(A) = $$\frac{1}{5}$$
∴ P(A) = $$\frac{1}{5}$$ × 2 = $$\frac{2}{5}$$

Question 28.
1998 tickets out of 2000 tickets do not have a prize. If a person randomly selects one ticket from 2000 tickets, then what is the probability that the ticket selected is eligible for prize ?
Total number of tickets = 2000
No. of tickets do not have a prize = 1998
No. of tickets eligible for prize = 2000 – 1998
= 2
Total number of outcomes of selecting a ticket is n = 2000c1 = 2000
A = Event that the selected ticket is eligible for prize
∴ m = 2c1 = 2
∴ P(A) = $$\frac{m}{n}$$
= $$\frac{2}{2000}$$
= $$\frac{1}{1000}$$

Section C

Answer the following questions as required:

Question 1.
Define the following events and draw their Venn diagram:
1. Mutually exclusive events
2. Union of events
3. Intersection of events
4. Difference event
5. Exhaustive events
6. Complementary event
1. Mutually exclusive events: Suppose, U is a finite sample space. A and B are any two events of U. If A ∩ B = Φ, then A and B are mutually exclusive events.

2. Union of events: Suppose, U is a finite sample space. A and B are any two events of U. Event A occurs or event B occurs or events A and B occur togethers, i.e., at least one of the events A and B occurs is called the union of events A and B. It is denoted by A ∪ B. Thus,
A ∪ B = {x; x ∈ A or x ∈ B or x ∈ A ∩ B}

3. Intersection of Events: Suppose, U is a finite sample space. A and B are any two events of U. The event that A and B occur together is called the intersection of events A and B. It is denoted by A ∩ B. Thus,
A ∩ B = {x; x ∈ A and x ∈ B}

4. Difference event: Suppose, U is a finite sample space. A and B are any two events of U. The event that A occurs but B does not occur is called the difference event of A and B. It is denoted by A – B or A ∩ B’. Similary, the event that A does not occur but B occurs is called the difference event of B and A. It is denoted by B – A or A’ ∩ B. Thus,
A – B = {x; x ∈ A and x ∉ B}
B – A = {x; x ∈ B and x ∉ A}

5. Exhaustive events: Suppose, U is a finite sample space. A and B are any two events of U. If A ∪ B = U, then A and B are called exhaustive events.

6. Complementary event: Suppose, A is any event of the finite sample space U. The event that A does not occur means the event consists of elements in U but not in A is called the complementary event of A. It is denoted by A’. Thus,
A’ = {x; x ∉ A, x ∈ U}

Question 2.
Give the illustrations of finite and infinite sample space.
Finite sample space: The sample space obtained for the random experiment throwing a balanced’ six face die U = {1, 2, 3, 4, 5, 6} is the illustration of finite sample space.

Infinite sample space: The sample space obtained for the random experiment of selecting a card from a pack of 52 cards till it is ace of heart is the illustration of infinite sample space.

Question 3.
Give the illustrations of impossible and certain event.
Impossible event: Event to get the number greater than 6 on the upper side of a balanced die.
Certain event: Event to get head or tail in tossing a balanced coin.

Question 4.
State the characteristics of random experiment.
The characteristics of random experiment are as follows :

• It can be repeated under identical conditions,
• Its all possible outcomes are known,
• It cannot be predicted with certainty which outcome will appear and
• It results into a certain outcome.

Question 5.
State the assumptions of mathematical definition of probability.
The assumptions of mathematical definition of probability are as follows:

• The number of outcomes in the sample space is finite.
• The number of all possible outcomes of the sample space is known.
• The outcomes of the sample space are equi-probable.

Question 6.
State the limitations of mathematical definition of probability.
The limitations of mathematical definition of probability are as follows:

• The probability of an event cannot be found if the outcomes are infinite.
• If the total number of outcomes is not known, the probability of an event cannot be found.
• If the elementary outcomes in the sample space are not equi-probable, the probability of an event cannot be found.

Question 7.
State the limitations of statistical definition of probability.
The limitations of statistical definition of probability are as follows:

• The infinite value of n cannot be taken in practice.
• The exact value of probability cannot be known.

Question 8.
Explain the equi-probable events with illustration.
If there is no apparant reason to believe that out of one or more events of a random experiment, any one event is more or less likely to occur than the other events, then those events are called equi-probable.

Illustration: In the random experiment of tossing a balance coin, two events of getting head (H) and getting tail (T) are equi-probable, because P (H) = P (T) = $$\frac{1}{2}$$

Question 9.
State the law of addition of probability for two events A and B. Write the law of addition of probability if these two events are mutually exclusive.
The law of addition of probability for two events A and B is as follows :
P(A ∪ B) = P(A) + P(B) – P(A ∩ B)
If A and B are mutually exclusive events, A ∩ B = 0 and P (A ∩ B) = 0. Hence, the law of addition of probability for two events A and B is written as follows:
P(A ∪ B) = P(A) + P(B)

Question 10.
State the law of multiplication of probability for two events A and B. Write the law of multiplication of probability if these two events are independent.
The law of multiplication of probability for two events A and B is as follows:
P(A ∩ B) = P(A|B) ∙ P(B) OR
P(A ∩ B) = P(B|A) ∙ P(A)
If two events A and B are independent, P(A|B) = P(A) and P(B|A) = P(B). So the law of multiplication of probability for two events A and B is written as follows:
P(AnB) = P(A) ∙ P(B)

Question 11.
State the following results for two independent events A and B:
1. P(A ∩ B)
2. P (A’ ∩ B’)
3. P(A ∩ B’)
4. P(A’ ∩ B)
For two independent events A and B, the results are obtained as follows:
1. P(A ∩ B) = P(A) × P(B)
2. P(A’ ∩ B’) = P(A’) × P(B’)
3. P(A ∩ B’) = P(A) × P(B’)
4. P(A’ ∩ B) = P(A’) × P(B) .

Question 12.
If p(A) = $$\frac{1}{3}$$, P(B) = $$\frac{2}{3}$$ and P(A ∩ B) = $$\frac{1}{6}$$, then find P(A’ ∩ B’).
p(A) = $$\frac{1}{3}$$, P(B) = $$\frac{2}{3}$$ and P(A ∩ B) = $$\frac{1}{6}$$ are given.
According to the law of addition of probability,
P(A ∪ B) = P(A) + P(B) – P(A ∩ B)
= $$\frac{1}{3}+\frac{2}{3}-\frac{1}{6}$$
= $$\frac{2+4-1}{6}$$
= $$\frac{5}{6}$$
Now, P(A’ ∩ B’) = P(A ∪ B)’
= 1 – P(A ∪ B)
= 1 – $$\frac{5}{6}$$
= $$\frac{6-5}{6}$$
= $$\frac{1}{6}$$

Question 13.
If P(B) = 2P(A|B) = 0.4, then find P(AnB).
Ans.
P(B) = 2P(A|B) = 0.4
∴ P(B) = 0.4, 2P(A|B) = 0.4
∴ P(A|B) = $$\frac{0.4}{2}$$ = 0.2
Now, P(A|B) = $$\frac{P(A \cap B)}{P(B)}$$
∴ P(A ∩ B) = P(A|B) ∙ P(B)
= 0.2 × 0.4 = 0.08

Question 14.
If the events A and B are independent and 3P(A) = 2P(B) = 0.12, then find P(A ∩ B).
3P(A) = 2P(B) = 0.12
∴ 3P(A) = 0.12 and 2P(B) = 0.12
∴ P(A) = $$\frac{0.12}{3}$$
= 0.04
and
P(B) = $$\frac{0.12}{2}$$
= 0.06
Now, A and B are independent events.
∴ P(A ∩ B) = P(A) – P(B)
= 0.04 × 0.06
= 0.0024

Question 15
If 5P (A) = 3P (B) = 2P (A ∪ B) = f for two events A and B, then find P(A’ ∪ B’).
5P (A) = 3P(B) = 2P (A ∪ B) = $$\frac{3}{2}$$
∴ 5P(A) = $$\frac{3}{2}$$ and 3P(B) = $$\frac{3}{2}$$ and 2P(A ∪ B) = $$\frac{3}{2}$$
∴ P(A) = $$\frac{3}{2 \times 5}$$ = $$\frac{3}{10}$$
∴ P(B) = $$\frac{3}{2 \times 3}$$ = $$\frac{1}{2}$$
∴ P(A ∪ B) = $$\frac{3}{2 \times 2}$$ = $$\frac{3}{4}$$
According to the law of addition of probability,
P(A ∪ B) = P(A) + P(B) – P(A ∩ B)
$$\frac{3}{4}$$ = $$\frac{3}{10}+\frac{1}{2}$$ – P(A ∩ B)
∴ P(A ∩ B) = $$\frac{3}{10}+\frac{1}{2}-\frac{3}{4}$$
= $$\frac{6+10-15}{20}$$ = $$\frac{1}{20}$$
Now, P(A’ ∪ B’) = P(A ∩ B)’
= 1 – P(A ∩ B) = 1 – $$\frac{1}{20}$$ = $$\frac{19}{20}$$

Question 16.
If P(A ∩ B) = 0.12 and P(B) = 0.3 for two independent events A and B, then find P(AuB).
A and B are independent events.
∴ P(A ∩ B) = P(A) – P(B)
∴ 0.12 = P(A) × 0.3
∴ P(A) = $$\frac{0.12}{0.3}$$ = 0.4
Now, according to the law of addition of probability,
P(A ∪ B) = P(A) + P(B) – P(A ∩ B)
= 0.4 + 0.3 – 0.12
= 0.58

Question 17.
If A = {x| 1 < x < 3} and B = {x|$$\frac{1}{2}$$ ≤ x < 2}, then find A ∪ B and A ∩ B.
A = {x| 1 < x < 3} and B = {x|-|- =£ x < 2}

From the above diagram, the events are obtained as follows:
(1) A ∪ B = {x|$$\frac{1}{2}$$ ≤ x < 3}
(2) A ∩ B = {x|1 < x < 2}

Question 18.
The probability of occurrence of at least one of the two events A and B is The probability that event A occurs, but event B does not occur is Find the probability of event B.
A ∪ B = Event that at least one of the two events A and B occurs
∴ P(A ∪ B) = $$\frac{1}{4}$$
AnB’ = Event that A occurs, but B does not occur
∴ P(A ∩ B’) = $$\frac{1}{5}$$
According to the law of addition of probability,
P(A ∪ B) = P(A) + P(B) – P(A ∩ B)
∴ P(A ∪ B) = P(A) – P(A ∩ B) + P(B)
[Putting, P(A ∩ B’) = P(A) – P(A n B)]
∴ P(A ∪ B) = P(A ∩ B’) + P(B)
∴ P(B) = P(A ∪ B) – P(A ∩ B’)
= $$\frac{1}{4}-\frac{1}{5}$$
= $$\frac{5-4}{20}$$
= $$\frac{1}{20}$$

Question 19.
If P(B) = $$\frac{3}{5}$$ and P(A’ ∩ B) = $$\frac{1}{2}$$ for two events A and B, find P(A|B).
P(B) = $$\frac{3}{5}$$ and P(A’ ∩ B) = $$\frac{1}{2}$$ are given.
∴ P(A’ ∩ B) = P(B) – P(A ∩ B)
∴ $$\frac{1}{2}$$ = $$\frac{3}{5}$$ – P(A ∩ B)
∴ P(A ∩ B) = $$\frac{3}{5}$$ – $$\frac{1}{2}$$ = $$\frac{6-5}{10}$$ = $$\frac{1}{10}$$
Now, P(A|B) = $$\frac{P(A \cap B)}{P(B)}$$
= $$\frac{1}{\frac{10}{\frac{3}{5}}}$$ = $$\frac{1}{10} \times \frac{5}{3}$$ = $$\frac{1}{6}$$
∴ P(A|B) = $$\frac{1}{6}$$

Question 20.
6 persons have a passport in’ a group of 10 persons. If 3 persons are randomly selected from this group, find the probability that
(i) all the three persons have a passport
(ii) two persons among them do not have a passport.
Out of 10 persons, 6 persons have a passport. So 4 persons do not have a passport.
Now, 3 persons out of 10 persons are randomly selected.
∴ Total number of primary outcomes,
n = 10C3 = $$\frac{10 \times 9 \times 8}{3 \times 2 \times 1}$$ = 120

(i) A = Event that all three persons have a passport
∴ Favourable outcomes for the event A is
m = 6C3 = $$\frac{6 \times 5 \times 4}{3 \times 2 \times 1}$$ = 120
P (A) = $$\frac{m}{n}$$ = $$\frac{20}{120}$$ = $$\frac{1}{6}$$

(ii) B = Event that two persons among three persons do not have a passport
∴ Favourable outcomes for the event B is
m = 6C1 × 4C2 = 6 × 6 = 36
∴ P(B) = $$\frac{m}{n}$$ = $$\frac{36}{120}$$ = $$\frac{3}{10}$$

Question 21.
The probability that the tax-limit for income of males increases in the budget of a year is 0.66 and the probability that the tax- limit increases for income of females is 0.72. The probability that the tax-limit increases for income of both the males and females is 0.47. Find the probability that
(i) the tax-limit increases for income of only one of the two, males and females,
(ii) the tax-limit does not increase for income of males as well as females in the budget of that year.
A = Event that the tax-limit for income of males increases.
B = Event that the tax-limit for income of females increases.
A ∩ B = Event that the tax-limit for income of male and female increases.
Here, P(A) = 0.66, P(B) = 0.72 and P(A ∩ B) =0.47 are given.

(i) C = Event that the tax-limit increases only for one of the two male and female.
Event C occurs in following two ways:
→ A ∩ B’ = Event that the tax-limit increases for the income of male only and not of female OR
→ A’ ∩ B = Event that the tax-limit increases for the income of female only and not of male Events A ∩ B’ and A’ ∩ B are mutually exclusive,
∴ P(C) = P(A ∩ B’) + P(A’ ∩ B)
= [P(A) – P(A ∩ B)l + (P(B) – P(A ∩ B)]
= [0.66 – 0.47] + [0.72 – 0.47]
= 0.19 + 0.25 = 0.44

(ii) A’ ∩ B’ = Event that the tax-limit does not increase for income of either male or female.
Now, P(A ∪ B) = P(A) + P(B) – P(A ∩ B)
= 0.66 + 0.72 – 0.47 = 0.91
∴ P(A’ ∩ B’) = P(A ∪ B)’ = 1 – P(A ∪ B)
= 1 – 0.91 = 0.09

Question 22.
The price of petrol rises in 80 % of the cases and the price of diesel rises in 77% of the cases after the rise in price of crude oil. The price of petrol and diesel rises in 68 % cases. Find the probability that the price of diesel rises under the condition that there is a rise in the price of petrol.
A = Event that the price of petrol rises
∴ P(A) = $$\frac{80}{100}$$
B = Event that the price of diesel rises
∴ P(B) = $$\frac{77}{100}$$
A ∩ B = Event that the prices of both petrol and diesel rises
∴ P(A ∩ B) = $$\frac{68}{100}$$
B|A = Event that the price of diesel rises knowing that the price of petrol rises
∴ P(B|A) = $$\frac{P(A \cap B)}{P(A)}$$
= $$\frac{\frac{68}{100}}{\frac{80}{100}}$$
= $$\frac{68}{100} \times \frac{100}{80}$$
= $$\frac{17}{20}$$

Question 23.
As per the prediction of weather bureau, the probabilities for rains on three days; Thursday, Friday and Saturday in the next week are 0.8, 0.7 and 0.6 respectively. Find the probability that it rains on at least one of the three days in the next week.
(Note:The events of rains on three days; Thursday, Friday and Saturday of a week are independent.)
A = Event that it rains on Thursday
B = Event that it rains on Friday
C = Event that it rains on Saturday
Here, P(A) = 0.8, P(B) = 0.7 and P(C) = 0.6 are given.
A, B and C are independent events.
∴ P(A ∩B) = P(A) ∙ P(B) = 0.8 × 0.7 = 0.56
P(A ∩ C) = P(A) ∙ P(C) = 0.8 × 0.6 = 0.48
P(B ∩ C) = P(B) ∙ P(C) = 0.7 × 0.6 = 0.42
P(A ∩ B ∩ C) = P(A) ∙ P(B) ∙ P(C)
= 0.8 × 0.7 × 0.6 = 0.336
Now, A ∪ B ∪ C = Event that it rains on at least one of the three days in the next week
According to the law of addition of probability,
P(A ∪ B ∪ C) = P(A) + P(B) + P(C) – P(A ∩ B) – P(A ∩ C) – P(B ∩ C) + P(A ∩ B ∩ C)
= 0.8 + 0.7 + 0.6 – 0.56 – 0.48 – 0.42 + 0.336
= 0.976

Section D

Question 1.
6 LED televisions and 4 LCD televisions are displayed in digital store A whereas 5 LED televisions and 3 LCD televisions are displayed in digital store B. One of the two stores is randomly selected and one television is selected from that store. Find the probability that it is an LCD television.
Digital store A: 6 LED TV + 4 LCD
TV = 10 TVs are displayed

Digital store A: 6 LED TV + 4 LCD
TV = 10 TVs are displayed

A1 = Event that a store is selected from the store A and B
∴ p(A1) = $$\frac{1}{2}$$

A2 = Event that LCD TV is selected from the store A
∴ p(A2) = $$\frac{{ }^{4} C_{1}}{{ }^{10} C_{1}}$$ = $$\frac{4}{10}$$

B1 = Event that if store A is selected then LCD TV is selected.
∴ p(B1) = p(A1) ∙ p(A2) = $$\frac{1}{2} \times \frac{4}{10}$$ = $$\frac{4}{20}$$

B2 = Event that if store B is selected then LCD TV is selected
∴ p(B2) = p(A1) ∙ p(A3) = $$\frac{1}{2} \times \frac{3}{8}$$ = $$\frac{3}{16}$$

C1 = Event that if one of the two stores is selected then LCD TV Is selected
= B1 ∪ B2
∴ P(C1) = P(B1) + P(B2)
= $$\frac{4}{20}+\frac{3}{16}$$
= $$\frac{16+15}{80}$$
= $$\frac{31}{80}$$

Question 2.
One number is randomly selected from the natural numbers 1 to 100. Find the probability that the number selected is either a single digit number or a perfect square.
One number is randomly selected from the numbers 1 to 100.
∴ Total number of primary outcomes,
n = 100C1 = 100
A = Event that the number selected is a single digit number
= {1. 2,3, 4,5,6,78, 9}
∴ m = 9
∴ P(A) = $$\frac{m}{n}$$ = $$\frac{9}{100}$$
B = Event that the number selected is a perfect square
= {1, 4, 9, 16, 25, 36, 49, 64, 81, 100)
∴ m = 10
∴ P(B) = $$\frac{m}{n}$$ = $$\frac{10}{100}$$
A ∩ B = Event that the selected number is a single digit number and a perfect square
= {1, 4,9}
∴ m = 3
∴ P(A ∩ B) = $$\frac{m}{n}$$ = $$\frac{3}{100}$$
Now, A ∪ B = Event that the number selected is either a single digit number or a perfect square
∴ P(A ∪ B) = P(A) + P(B) – P(A ∩ B)
= $$\frac{9}{100}+\frac{10}{100}-\frac{3}{100}$$
= $$\frac{16}{100}$$
= $$\frac{4}{25}$$

Question 3.
A balanced coin is tossed thrice. If the first two tosses have resulted in tail, find the probability that tail appears on the coin In all the three trials.
A = Event that in the first trial tall is obtained.
B = Event that in the second trial tail is obtained on the coin.
∴ P(A) = $$\frac{1}{2}$$, P(B) = $$\frac{1}{2}$$
C = Event that in the first two trials tail is obtained on the coin.
∴ P(C) = P(A) ∙ P(B)
= $$\frac{1}{2} \times \frac{1}{2}$$
= $$\frac{1}{4}$$
D|C = Event that tall appears on the in all three trials if the first two trials resulted in tail
∴ P(D|C) = $$\frac{P(D \cap C)}{P(C)}$$
D ∩ C = Event that in the first two trails and In the third trial tail is obtained on the coin
∴ P(D ∩ C) = P(D) ∙ P(C)
= $$\frac{1}{2} \times \frac{1}{4}$$
= $$\frac{1}{8}$$
Now, P(D|C) = $$\frac{P(D \cap C)}{P(C)}$$ = $$\frac{\frac{1}{8}}{\frac{1}{4}}$$
= $$\frac{1}{8} \times \frac{4}{1}$$
= $$\frac{1}{2}$$

Question 4.
If events A, B and C are independent events and P(A) = P(B) = P(C) = p, then find the value of P(A ∪ B ∪ C) in terms of p.
Events A, B and C are independent events.
P(A) = P(B) = P(C) = p
P(A ∪ B ∪ C) = P(A) + P(B) + P(C) – P (A ∩ B) – P (A ∩ C) – P(B ∩ C) + P(A ∩ B ∩ C)
= P(A) + P(B) + P(C) – P(A) ∙ P(B) – P(A) ∙ P(C) – P(B) ∙ P(C) + P(A) ∙ P(B) ∙ P(C)
= p + p + p – (p × p) – (p × p) – (p × p) + (p × p × p)
= 3p – p2 – p2 – p2 + p3
= 3p – 3p2 + p3
= p (3 – 3p + p2)

Question 5.
The genderwise data of a sample of 6000 employees selected from class 3 and class 4 employees in the government jobs of a state are shown in the following table:

One employee is randomly selected from all the class 3 and class 4 employees in government jobs of this state.
(1) If the selected employee is a male, find the probability that he belongs to class 3.
(2) If it is given that the selected employee belongs to class 3, find the probability that he is a male.
Suppose, A = Event that an employee belongs to class 3
B = Event that an employee belongs to class 4
C = Event that an employee is a male
D = Event that an employee is a female
∴ P(A) = $$\frac{4500}{6000}$$, P(B) = $$\frac{1500}{6000}$$, P(C) = $$\frac{4000}{6000}$$ and P(D) = $$\frac{2000}{6000}$$

(1) A|C = Event that employee selected is a male, then he belongs to class 3
P(A|C) = $$\frac{P(A \cap C)}{P(C)}$$
= $$\frac{\frac{3600}{6000}}{\frac{4000}{6000}}$$
= $$\frac{3600}{4000}$$
= $$\frac{9}{10}$$

(ii) C|A = Event that an employee selected belongs to class 3, then he is a male.
P(C|A) = $$\frac{P(A \cap C)}{P(C)}$$
= $$\frac{\frac{3600}{6000}}{\frac{4500}{6000}}$$
= $$\frac{3600}{4500}$$
= $$\frac{4}{5}$$