1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Conditional probability with urns

  1. Oct 6, 2012 #1


    User Avatar
    Gold Member

    1. The problem statement, all variables and given/known data
    Consider 3 urns. Urn A contains 2 white and 4 red balls, Urn B contains 8 white and 4 red balls and Urn C contains 1 white and 3 red balls. If 1 ball is selected from each urn, what is the probability that the ball chosen from urn A was white given that exactly 2 white balls were selected?

    3. The attempt at a solution
    Let E be the event that a white ball was chosen from urn A
    Let F be the event that exactly 2 white balls were selected.

    Given that the same colour of balls are indistinguishable , we have |s| = (2 choose 1) x (2 choose 1) x (2 choose 1) = 8 possibilities
    Therefore F = {(WRW),(WWR),(RWW)} and E = {(WWW),(WRW),(WWR),(WRR)}
    So, [tex] P(E|F) = \frac{P(EF)}{P(F)} = \frac{\frac{2}{8}}{\frac{3}{8}} = 2/3 [/tex]
    The given answer is 7/11, which is slightly below this - where is the error?
    Many thanks.
  2. jcsd
  3. Oct 6, 2012 #2


    User Avatar
    Science Advisor

    Your basic concept is correct but you appear to have done the arithmetic incorrectly- at least your "P(EF)= 2/9" and "P(F)= 3/8" are wrong but you don't show how you got those.

    Yes, P(F), the probability that exactly two white balls are drawn can be calculated by looking at "WRW", "WWR", and "RWW". Since the first urn contains two white balls and four red, the probability that the first ball drawn is white is 2/6= 1/3 and the probability that it is red is 4/6= 2/3. The second urn contains eight white and four red balls so that the probability the second ball drawn is white is 8/12= 2/3 and the probability it is red is 4/12= 1/3. The third urn contains one white and three red balls so the probability the third ball drawn is white is 1/4 and the probability it is red is 3/4.

    So the probability of "WRW" is (1/3)(1/3)(1/4)= 1/36. The probability of "WWR" is (1/3)(2/3)(3/4)= 1/6. The probability of "RWW" is (2/3)(2/3)(1/4)= 1/9. The probability of "two white", in any order, is P(F)= 1/36+ 1/6+ 1/9= 1/36+ 6/36+ 4/36= 11/36, not "3/8".

    The event "the first ball drawn is white and two white balls are drawn" just drops "RWW" from the possible cases so the probability of that is 1/36+ 1/6= 1/36+ 6/36= 7/36.
  4. Oct 6, 2012 #3


    User Avatar
    Gold Member

    I said EF = {(WRW),(WWR)} and the probability of attaining this out of the sample space was 2/8. Similarly, F = {(WRW),(WWR),(RWW)} which gives P(F) = 3/8. (3 elements out of sample space of 8 elements)
    This method does not take into account the numbers of red and white balls in ech urn, but at the same time, I don't see anything wrong with what I have done.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook