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Math Help - Conditional Probability Brain Teaser

  1. #1
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    Conditional Probability Brain Teaser

    A sock contains 2 red marbles and an unknown number of blue marbles. Tom places a new marble in the sock. Jason puts his hand in the sock and pulls out a red marble.

    What is the probability that the marble Tom put in the sock was red?


    I've been trying to approach this with Bayes' Theorem but not making any progress:

    P(A l B) = P(B l A) * P(A) / P(B)

    I'm confused as to what A and B would represent in the problem.

    Appreciate the help, thanks!
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  2. #2
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    Quote Originally Posted by Penguins View Post
    A sock contains 2 red marbles and an unknown number of blue marbles. Tom places a new marble in the sock. Jason puts his hand in the sock and pulls out a red marble.

    What is the probability that the marble Tom put in the sock was red?


    I've been trying to approach this with Bayes' Theorem but not making any progress:

    P(A l B) = P(B l A) * P(A) / P(B)

    I'm confused as to what A and B would represent in the problem.

    Appreciate the help, thanks!
    Let the number of blue marbles in the sock be x.

    Draw a tree diagram. The first two branches are the events Tom putting in a blue marble or a red marble. The next branches are the events of pulling out a red marble or a blue marble.

    I get \frac{\frac{3}{2(3 + x)}}{\frac{3}{2(3 + x)} + \frac{1}{3 + x}} = \frac{3}{5}.
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  3. #3
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    I don't understand how you set up the problem like you did. What do the terms in your numerator and denominator represent?

    Here is what I get:

    P(A l B) = (P(B l A) * P(A)) / P(B)

    A is Tom putting a red marble in the sock.
    B is Jason taking a red marble out of the sock.

    P (B l A) = 3 / (x+3)
    P (A) = unknown, let it be p
    P (B) = p*(3/(x+3)) + (1-p)*(2/(x+2))

    So

    P(A l B) = (P(B l A) * P(A)) / P(B) = (3p/(x+3)) / (p*(3/(x+3)) + (1-p)*(2/(x+2)))

    I end up with 3/2 p. I know I did something wrong, but I'm not sure what.
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  4. #4
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    Hello, Penguins!

    I agree with Mr. F


    A sock contains 2 red marbles and an unknown number of blue marbles.
    Tom places a new marble in the sock.
    Jason puts his hand in the sock and pulls out a red marble.

    What is the probability that the marble Tom put in the sock was red?
    We want: . P(\text{add Red }|\text{ draw Red}) \;=\;\frac{P(\text{add Red }\wedge\text{ draw Red})}{P(\text{draw Red})} .[1]

    The sock contains: 2 Red and b Blue marbles.


    There are two cases to consider:

    [1] A Red marble is added.
    . . .The sock contains: 3 Red and b Blue.
    . . . . Then: . P(\text{add Red }\wedge\text{ draw Red}) \:=\:\frac{3}{b+3} .[2]

    [2] A Blue marble is added.
    . . .The sock contains: 2 Red and b+1 Blue.
    . . . . Then: . P(\text{add Blue }\wedge\text{ draw Red}) \:=\:\frac{2}{b+3}

    Hence: . P(\text{draw Red}) \;=\;\frac{3}{b+3} + \frac{2}{b+3} \;=\;\frac{5}{b+3} .[3]


    Substitute [2] and [3] into [1]: . P(\text{add Red }|\text{ draw Red}) \;\;=\;\; \frac{\dfrac{3}{b+3}} {\dfrac{5}{b+3}} \;\;=\;\;\frac{3}{5}


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  5. #5
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    Quote Originally Posted by Soroban View Post
    Hello, Penguins!

    I agree with Mr. F


    We want: . P(\text{add Red }|\text{ draw Red}) \;=\;\frac{P(\text{add Red }\wedge\text{ draw Red})}{P(\text{draw Red})} .[1]

    The sock contains: 2 Red and b Blue marbles.


    There are two cases to consider:

    [1] A Red marble is added.
    . . .The sock contains: 3 Red and b Blue.
    . . . . Then: . P(\text{add Red }\wedge\text{ draw Red}) \:=\:\frac{3}{b+3} .[2]

    [2] A Blue marble is added.
    . . .The sock contains: 2 Red and b+1 Blue.
    . . . . Then: . P(\text{add Blue }\wedge\text{ draw Red}) \:=\:\frac{2}{b+3}

    Hence: . P(\text{draw Red}) \;=\;\frac{3}{b+3} + \frac{2}{b+3} \;=\;\frac{5}{b+3} .[3]


    Substitute [2] and [3] into [1]: . P(\text{add Red }|\text{ draw Red}) \;\;=\;\; \frac{\dfrac{3}{b+3}} {\dfrac{5}{b+3}} \;\;=\;\;\frac{3}{5}



    I see what you're doing, but isn't it a problem that you're assuming the probability of Tom putting a red marble in the sock to be 1/2? I set up the problem using p as this probability and unfortunately made a careless mistake in my previous post, but get the answer to the question as (3p)/(2+p). When p = 1/2, the answer is 3/5.

    I'm curious as to whether there is a rule in a situation like this as to whether lacking more information it is assumed that all possibilities have equal likelihood?


    Thanks for the replies!
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  6. #6
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    Quote Originally Posted by Penguins View Post
    I see what you're doing, but isn't it a problem that you're assuming the probability of Tom putting a red marble in the sock to be 1/2? I set up the problem using p as this probability and unfortunately made a careless mistake in my previous post, but get the answer to the question as (3p)/(2+p). When p = 1/2, the answer is 3/5.

    I'm curious as to whether there is a rule in a situation like this as to whether lacking more information it is assumed that all possibilities have equal likelihood?


    Thanks for the replies!
    If you're expected to get a numerical answer then you have to make an assumption regarding a priori knowledge. The best one (for various reasons which I won't get into here) is a uniform prior distribution.
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