Question

Consider 3 urns. Urn A contains 2 white and 4 red balls, urn B contains 8 white and 4 red balls, and urn C contains 1 white and 3 red balls. If 1 ball is selected from each urn, what is the probability that the ball chosen from urn A was white given that exactly 2 white balls were selected?

Answer 1

Answer:0.636

Explanation:

Given

Urn A contains 2 W, 4 R

Urn B contains 8 W, 4 R

Urn C contains 1 W,  3 R

Probability of selecting 3 balls out of which 2 are white

P(J)=P(WWR)+P(WRW)+P(RWW)

$P(J)=\frac{2}{6}\cdot \frac{8}{12}\cdot \frac{3}{4}+\frac{2}{6}\cdot \frac{4}{12}\cdot \frac{1}{4}+\frac{4}{6}\cdot \frac{8}{12}\cdot \frac{1}{4}=\frac{88}{288}$

Probability that the ball chosen from Urn A was white given that exactly 2 white balls  were selected

i.e.

$P(A/J)=\frac{P\left ( A\cap J\right )}{P\left ( J\right )}$

$P(A/J)=\frac{\frac{56}{288}}{\frac{88}{288}}=0.636$