Probability Problem Q14

If 1 ball is selected from each urn, what is the probabililty that the ball chosen from Urn A was white, given that exactly two white ball were selected?
Briefly, there are 3 situations for 2 white balls( and 1 red ball) :
(To help you understand, I use [white] [red] to denote white ball and red ball in urn respectively.)
(1) A[white]B[white]C[red]: P(1)=(2/6)(8/12)(3/4)=6/36
(2) A[white]B[red]C[white}: P(2)=(2/6)(4/12)(1/4)=1/36
(3) A[red]B[white}C{white}: P(3)=(4/6)(8/12)(1/4)=4/36
From above, firstly work out the following :
X : the probability that exactly two white ball were selected, when 1 ball was selected from each urn-- all the above situations (1), (2) & (3)
X = P(1) + P(2) + P(3) = 11/36
Y: the probability that white ball was selected from Urn A, and another white ball was selected from either Urn B or Urn C-- situations (1) & (2) above
Y = P(1) + P(2) = 7/36
Therefore 'the probabililty that the ball chosen from Urn A was white, given that exactly two white ball were selected' can be calculated as:
Y / X =( 7/36) / (11/36) = 7/11

Good

First find out the probability that exactly two white ball were selected

This probability is equal to

(2/6)(8/12)(3/4)+(4/6)(8/12)(1/4)+(2/6)(4/12)(1/4)

=88/288

=11/36

The ball chosen from urn a was white and exactly two white ball were selected

=(2/6)(8/12)(3/4)+(2/6)(4/12)(1/4)

=7/36

2009-01-05 16:08:27 補充：
So the probabililty that the ball chosen from urn a was white, given that exactly two white ball were selected

=7/11 (using conditional probability)

VERY GOOD ANSWER