box A contains 2 red balls,
box B contains 2 white balls,
box C contains 1 red ball and 1 white ball,
A box is selected at random (with equal probabilities) and one ball is taken at random from that box.
(i) Compute the probability of selecting a white ball using Bayes Rule.
A(1/3) _____ Red(1/6)
.........|____ Red(1/6)
B(1/3)______White(1/6)
........|_____White(1/6)
C(1/3)______Red(1/6)
........|_____White(1/6)
so.. P(White) = [(1/3)*(1/6)] + [(1/3)*(1/6)] + [(1/3)*(1/6)]
= 1/6
Correct?
(ii)Compute the conditional probability that B has been selected given that the ball is white.
P(B|White) = P(B intersect White) / P(White)
what is my P(B intersect White)?
No, P(W|A) represents the probability of selecting a white ball, given you are selecting from box A. Since there are no white balls in box A, the probability P(W|A)=0. So, following that train of thought we have
P(W|A)P(A)=(0/2)(1/3)=0
P(W|B)P(B)=(2/2)(1/3)=(1/3)
P(W|C)P(C)=(1/2)(1/3)=(1/6).
The below equation is from wikipedia:Bayes' theorem - Wikipedia, the free encyclopedia
In terms of your problem, you need to compute
(we are partitioning the boxes).
Notice you have already figured out both the numerator and denominator.