SOLUTION: Urn A contains seven white balls and five black balls. Urn B contains three white balls and six black balls. A ball is drawn from Urn A and then transferred to Urn B. A ball is the

Algebra ->  Probability-and-statistics -> SOLUTION: Urn A contains seven white balls and five black balls. Urn B contains three white balls and six black balls. A ball is drawn from Urn A and then transferred to Urn B. A ball is the      Log On


   

Question 1118549: Urn A contains seven white balls and five black balls. Urn B contains three white balls and six 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.)
Answer by greenestamps(13214) About Me  (Show Source):
You can put this solution on YOUR website!


Formally, the definition of conditional probability is

P(A|B) = (P(A and B))/(P(B))

The way I think of conditional probability is that the sample space is only the outcomes in which B happens, so P(B) is the denominator of the probability fraction. The numerator is then the probability that BOTH A and B happen.

In your problem, the probability of getting white and then white is

P%28WW%29+=+%287%2F12%29%2A%284%2F10%29+=+28%2F120

and the probability of getting black and then white is

P%28BW%29+=+%285%2F12%29%2A%283%2F10%29+=+15%2F120

Then the probability that the first ball was white, given that the second ball was white, is

%2828%2F120%29%2F%28%2828%2F120%29%2B%2815%2F120%29%29+=+28%2F%2828%2B15%29+=+28%2F43