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Understanding Conditional Probabilities

  1. Jun 26, 2012 #1
    1. The problem statement, all variables and given/known data
    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?
     
    Last edited: Jun 26, 2012
  2. jcsd
  3. Jun 26, 2012 #2
    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.
     
  4. Jun 27, 2012 #3

    HallsofIvy

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    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.
     
  5. Jun 27, 2012 #4
    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 [itex]\frac{\frac{3}{7}}{\frac{3}{7} + \frac{2}{3}} = \frac{9}{23}[/itex].

    Unless my logic is wrong.
     
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