# Homework Help: Independent Events Probability

1. Jan 29, 2012

### lina29

1. The problem statement, all variables and given/known data
During a winter season at one ski resort, there are two roads from Area A to Area B and two roads from Area B to Area C. Each of the four roads is blocked by snow with a probability p=.25 independently of the other roads.

What is the probability that there exists an open route from Area A to Area C

It was decided to add a new path connecting areas A and C directly. It is also blocked by snow with probability p=.25 independently of all the other paths. Now what is the probability that there is an open route from area A to area C.

2. Relevant equations

3. The attempt at a solution
For the first part I assumed that at least one road had to be open from A to B so I got .4375 which is the same for at least one road from B to C. And then that both A to B and B to C had to be open so I got .191 which was wrong.

Any help would be appreciated

2. Jan 29, 2012

### jimbobian

Well, I'm sure there is probably an easier way but there is nothing wrong with drawing a good old probability tree. It will be quite a big one (as there are four roads and so 4 "rounds" to the tree). But once you've drawn the tree, figure out which branches correspond to it being possible to get from A to C, then figure out their probabilities and you should get the answer.

James

3. Jan 29, 2012

### LCKurtz

Remember that the probability of a road between two points being open is 1 minus the probability that both are blocked.

4. Jan 29, 2012

### lina29

Right so what I did to find at least one road being open between A and B or B and C was
1-(1-.25)(1-.25)= .4375
and then to find the probability of both being open I did
.4375*.4375=.191

5. Jan 29, 2012

### jimbobian

But the probability of being blocked is .25, so you've worked out the opposite!

6. Jan 29, 2012

### lina29

Ohh so what I would do is 1-(1-.75)(1-.75)= .5
and then
.5*.5=.25 which would be the final answer for the first part?

7. Jan 29, 2012

### jimbobian

I agree with the logic, not the answer ;)

8. Jan 29, 2012

### lina29

sorry :)
it would be 1-(1-.75)(1-.75)= .9375
and then
.9375*.9375=.8789 right?

For the second part how would I approach it?

9. Jan 29, 2012

### jimbobian

Sounds good.

Well now you have the probability that there is a route to C through B that is open, and you also know the probability of getting directly from A to C. Can you think of a way of combining these to get the second answer?

10. Jan 29, 2012

### lina29

my though was addition but then the probabilities would be over 1

11. Jan 29, 2012

### jimbobian

Probabilities over 1 are never really a good sign!

Well imagine you've got two coins, what would be the probability of at least 1 head?

12. Jan 29, 2012

### lina29

1-(1-.5)(1-.5)=.75

13. Jan 29, 2012

### jimbobian

Good, so can you see what the probability of at least one route being open is?

14. Jan 29, 2012

### lina29

1-(1-.8789)(1-.75)=.9697

15. Jan 29, 2012

### jimbobian

Yep, I would agree

16. Jan 29, 2012

### lina29

thank you!

17. Jan 29, 2012

### jimbobian

No problem, hope they're right!

18. Jan 29, 2012

### HallsofIvy

That is correct but you did it the hard way. If P(A)= .25 and P(B)= .25, and A and B are independent, $P(A and B)= .25^2= 0.0625$ the probability of both roads being blocked is 0.0625 so the probability that at least one of the roads is not is $1- P(A)P(B)= 1- .25^2= 1- 0.0625= 0.9375$, as you say.

19. Jan 29, 2012

### Ray Vickson

I get something different from all of you!

In the first problem, {AC blocked} = {both AB blocked} or {both BC blocked}, so P{AC blocked} = (1/4)(1/4) + (1/4)(1/4) - (1/4)^4 = 31/256 = .12109375, so P{AC open} = 225/256 = .87890625 . This uses P{A or B} = P{A} + P{B} - P{A & B}.

RGV