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Probability Question - colored balls in 2 bowls - Baye's formula?

  1. Feb 4, 2013 #1
    Probability Question -- colored balls in 2 bowls -- Baye's formula?

    Bowl #1 contains 2 red balls and 2 white balls
    Bowl #2 contains 3 red balls and 2 white ball

    one bowl is chosen at random (each is equally likely)

    (A) What is the probability of choosing a red ball?

    Let R stand for choosing a red ball
    [itex]B_{#}[/itex] = Bowel #

    [itex]P(R) = P(B_{1})P(R|B_{1})+P(B_{2})P(R|B_{2})[/itex]
    [itex]P(R) = \frac{1}{2} \frac{1}{2} + \frac{1}{2} + \frac{3}{4} = .625[/itex]

    (B)
    [itex]P(B_{1}) = P(R)P(B_{1}|R)[/itex]

    I don't know how to calculate [itex]P(B_{1}|R)[/itex]. Do I need to use Baye's formula?

    Thanks for any help! Hopefully I'm not making this to complicated again lol.
     
    Last edited by a moderator: Feb 4, 2013
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  3. Feb 4, 2013 #2

    Dick

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    Re: Probability Question -- colored balls in 2 bowls -- Baye's formula?

    I'm not an expert in these things but you got part (A) wrong. Try it again. And for part (B), now it sounds like a good time to use Baye's formula.
     
    Last edited: Feb 4, 2013
  4. Feb 4, 2013 #3
    Re: Probability Question -- colored balls in 2 bowls -- Baye's formula?

    Is it just (1/2)(3/4) for part A?
     
  5. Feb 4, 2013 #4

    Dick

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    Re: Probability Question -- colored balls in 2 bowls -- Baye's formula?

    No, no. It's given by exactly the formula you wrote except you've got P(R|B_2) wrong. What is that? It's not 3/4.
     
  6. Feb 4, 2013 #5

    haruspex

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    Re: Probability Question -- colored balls in 2 bowls -- Baye's formula?

    What is the actual question for B?
    That would not be true in general. [itex]P(B_{1} \wedge R) = P(R)P(B_{1}|R)[/itex], [itex]P(B_{1}) = P(R)P(B_{1}|R) + P(\neg R)P(B_{1}|(\neg R) [/itex]
     
  7. Feb 4, 2013 #6
    Re: Probability Question -- colored balls in 2 bowls -- Baye's formula?

    Ok so I figured out my mathematical miscalculation in A and was able to figure out the answer to B (I realized I forgot to post it).

    What is that symbol mean?
    ¬?
     
  8. Feb 5, 2013 #7

    HallsofIvy

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    Re: Probability Question -- colored balls in 2 bowls -- Baye's formula?

    Surely you don't mean that second "+"?
    [tex]\frac{1}{2}\frac{1}{4}+ \frac{1}{2}\frac{3}{5}= \frac{1}{8}+ \frac{3}{10}[/tex]
    Ignoring your typo with the "+", the probability that, drawing from the second bowl, you will draw a red ball, is 3/(3+ 2)= 3/5, not 3/4.

    What, exactly, is the question? This is simply a "formula". What are you to do with it?
    (And, by the way, an incorrect formula. After having picked a single ball, the probability that it was picked from bowl 1 is [itex]P(B_1)= P(R)P(B_1|R)+ P(W)P(B_1|W)[/itex])

    That at least is easy to answer. There are a total of 5 red balls, 2 of them from bowl 1. Given that you picked a red ball, the probability it came from bowl 1 is 2/5.

     
  9. Feb 5, 2013 #8

    haruspex

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    Re: Probability Question -- colored balls in 2 bowls -- Baye's formula?

    It's a logical 'not'.
     
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