Question

Urn A contains four white balls and five 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.)

Answer 1

Answer:

If the seond ball drawn was white, there is a 48.3% probability that the transferred ball was white.

Step-by-step explanation:

This can be formulated as the following problem:

What is the probability of B happening, knowing that A has happened.

It can be calculated by the following formula

$P = \frac{P(B).P(A/B)}{P(A)}$

Where P(B) is the probability of B happening, P(A/B) is the probability of A happening knowing that B happened and P(A) is the probability of A happening.

For this problem, we have:

What is the probability of the transfered ball being white, given that the second ball drawn was white.

P(B) is the probability that the ball transfered is white. The ball is transferred from urn A to urn B. Urn A contains four white balls and five black balls. So $P(B) = \frac{4}{9} = 0.444$

P(A/B) is the probability that the second ball drawn is white, given that a white ball was transferred. If a white ball was transferred, there will be 10 balls in Urn B, 7 of which are white. So $P(A/B) = 0.7$.

P(A) is the probability that the ball drawn for the second urn is white. There are two cases. Either a black ball is transfered(with probability $\frac{5}{9} = 0.556$) and then a white ball is drawn, with probability $\frac{6}{10} = 0.6$. Or we have P(A/B), with probability 0.444. So

$P(A) = 0.444*0.7 + 0.556*0.6 = 0.644$

Finally

$P = \frac{P(B).P(A/B)}{P(A)} = \frac{0.444*0.7}{0.644} = 0.483$

If the seond ball drawn was white, there is a 48.3% probability that the transferred ball was white.