- #1
CmdrGuard
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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 [itex]P_{N}(R)[/itex], that is, the probability of picking the right pocket, given that you already picked a nickel.
I understand that
[itex]P(NR) = P(N)\bullet P_{N}(R)[/itex].
I figured that [itex]P(NR) = \frac{1}{2}\frac{3}{7}[/itex] 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 [itex]P(N)=\frac{5}{10}[/itex] since out of the total 10 coins in both pockets, 5 of them are nickels.
So I simply solved for [itex]P_{N}(R)[/itex] and I got [itex]\frac{3}{7}[/itex], which is wrong.
But where is my logic incorrect?
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