Question

Urn A contains four white balls and three black balls. Urn B contains six white balls and seven 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.)

I can't seem to figure this out! Please help! Thank you!

Answer 1

Let W = {white ball transferred from Urn A}

         B = {Black ball transferred from Urn A}

P(W) = 4/7

P(B) = 3/7

A be the event that the ball drawn from urn B is white then

P(A) = P(W)P(A|W) + P(B)P(A|B)

If white ball is transferred to Urn B then Urn B contains 7 white balls and 7 black balls i.e. 14 total balls

P(A|W) = 7/14

If black ball is transferred to Urn B then Urn B contains 6 white balls and 8 black balls i.e. 14 total balls

P(A|B) = 6/14

P(A) = P(W)P(A|W) + P(B)P(A|B)

          = 4/7 * 7/14 + 3/7 * 6/14

          = 23/49 = 0.4693

We want to find the probability that the transferred ball was white given that the second ball drawn was white.

Using Bayes' theoram,

P(W|A) = P(W) P(A|W) / P(A) = (4/7) * (7/14) / (23/49) =0.6086

The probability that the transferred ball was white given that the second ball drawn was white is 0.6086