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Conditional Probability Question

  1. Sep 17, 2014 #1
    1. The problem statement, all variables and given/known data
    Suppose there are two urns that contain white and yellow balls. Urn 1 contains
    10 white and 5 yellow balls, and Urn 2 contains 6 white and 12 yellow balls. You
    are going to draw 3 balls without replacement from one of the urns. To decide
    which urn to draw from, you will roll an ordinary six-sided (balanced) die. If the
    die comes up with a 1, you will draw 3 balls from Urn 1, otherwise you will draw
    3 balls from Urn 2.
    (a) What is the probability you draw exactly two white balls?
    (b) Given you draw exactly two white balls, what is the probability you rolled a
    1?


    2. Relevant equations
    (10,2) for instance would read ten choose 2 in combinatorics

    3. The attempt at a solution

    Ok, so I'm not entirely sure how to do this, but this is my attempt..

    Urn #1)
    10 white balls
    5 yellow

    ((10,2)(5,1))/(15,3) = P(choose exactly 2 white balls from urn #1) = P(#1) = 45/91

    Urn #2)

    6 white balls
    12 yellow

    ((6,2)(12,1))/(18,3) = P(choose exactly 2 white balls from urn #2) = P(#2) = 15/68

    -----

    P(draw from urn 1) = 1/6 = P(#11)
    P(draw from urn 2) = 5/6 = P(#22)

    P(draw exactly 2 white balls) = P(#1)P(#11) + P(#2)P(#22) = (45/91)(1/6) + (15/68)(5/6) = 0.26624

    PART B)

    P(roll a 1 | draw 2 white balls) = P(roll a 1 and draw 2 white balls)/P(draw 2 white balls)

    Am I on the right track here?
     
  2. jcsd
  3. Sep 18, 2014 #2

    BvU

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Wrong kind of question for this forum: PF is not in the business of stamp-approving homework. Wouldn't be good for our relations with the poor teachers of this world.

    Show some self-assurance !

    I have no clue what would be the right track, but I enjoyed following yours. Thanks for sharing this with us ! :approve:
     
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