Understanding Conditional Probabilities

[CmdrGuard]

Homework Statement

Suppose you have 3 nickels and 4 dimes in your right pocket and 2 nickels and a quarter in your left pocket. You pick a pocket at random and from it select a coin at random. If it is a nickel, what is the probability that it came from your right pocket?

2. The attempt at a solution

Let N be the event of picking a nickel, and R be the event of picking the right pocket.

My understanding is as follows:

What the question is asking for is P_{N}(R), that is, the probability of picking the right pocket, given that you already picked a nickel.

I understand that
P(NR) = P(N)\bullet P_{N}(R).

I figured that P(NR) = \frac{1}{2}\frac{3}{7} because there is a 50% chance I pick the right pocket and then a 3/7th chance that within that pocket I pick a nickel.

If I am making a mistake I suspect this is it.

Then I also figured that P(N)=\frac{5}{10} since out of the total 10 coins in both pockets, 5 of them are nickels.

So I simply solved for P_{N}(R) and I got \frac{3}{7}, which is wrong.

But where is my logic incorrect?

 

[Villyer]

P(N) looks wrong, you cannot say that there are 5 total nickels and 10 total coins. Instead, you have to sum the probability of picking a nickel out of the right pocket and picking a nickel out of the left pocket.

 

[HallsofIvy]

Yes, you can say "there are 5 total nickels and 10 total coins". That is given. What you cannot do is say "there is a 50% chance I pick the right pocket" when that is the probability you are asked to find.

You have a total of 5 nickels, three in your left pocket and two in your right. If you take a nickel out of your pockets, the probability it came from your right pocket is 2/5.

 

[Villyer]

Yes, you can say "there are 5 total nickels and 10 total coins". That is given.

You're right, what I meant to say was that you cannot use that fact alone to find his P(N), you need to know the distribution in each pocket.But the answer I got was 9/23.
There is a 3/7 chance of picking a nickel out of the right pocket, and a 2/3 chance of picking a nickel out of the left pocket. So the probability of the nickel coming from the right pocket is \frac{\frac{3}{7}}{\frac{3}{7} + \frac{2}{3}} = \frac{9}{23}.

Unless my logic is wrong.