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)
[tex]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}[/tex]
Probability that the ball chosen from Urn A was white given that exactly 2 white balls were selected
i.e.
[tex]P(A/J)=\frac{P\left ( A\cap J\right )}{P\left ( J\right )}[/tex]
[tex]P(A/J)=\frac{\frac{56}{288}}{\frac{88}{288}}=0.636[/tex]
