Conceptually this is the same category of problems as the previous boomers problem.
Let events
w=a white ball was transferred
b=a black ball was transferred
now concentrate on urn B.
W=a white ball was drawn
B=a black ball was drawn
then we know that
probability of transferring a white ball
P(w)=4/11
and
P(b)=7/11.
If a white ball was transferred, then probability of drawing a white ball from urn B
P(W)=7/10, or
P(W|w)=7/10
and similarly,
P(W|b)=6/10
Applying the law of total probability
P(W)=P(W|w)*P(w)+P(W|b)*P(b)
=(7/10)(4/11)+(6/10)*(7/11)
=7/11
(appreciate the advantages of working with fractions in probability calculations)
Applying Bayes theorem
P(w|W)=P(W|w)*P(w)/P(W)
=(7/10)*(4/11)/(7/11)
=(28/110)/(7/11)
=2/5
Urn A contains four white balls and seven black balls. Urn B contains six white balls and three black balls. A ball is drawn from Urn A and then transferred to Urn B. A ball is then drawn from Urn B. What is the probability that the transferred ball was white given that the second ball drawn was white? (Round your answer to three decimal places.)
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