Q.3.9 - A First Course in Probability

Question

Consider$3$urns. An urn $A$ contains$2$ white and $4$ red balls, an 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?

Step-by-Step Solution

Answer

$0.636365$is the probability that the ball chosen from urn $A$ was white given that exactly two white balls were selected.

Step 1:Given Information

Given that $3$ urns. An urn $A$ contains 2 white and 4 red balls, an urn $B$ contains eight white and four red balls, and an urn $C$contains one white and three red balls.

Step 2:Explanation

Let $X$ be the possibility that exactly two White balls are selected i.e.,$(W,W,R)$$(W,R,W) (R,W,W)$

$\mathrm{p}\left(x\right)=\frac{2}{6}\times \frac{8}{12}\times \frac{3}{4}+\frac{2}{6}\times \frac{4}{12}\times \frac{1}{4}+\frac{4}{6}\times \frac{8}{12}\times \frac{1}{4}$

Let $Y$ be the possibility that the first ball is white i.e.,$(W,W,R)$$(W,R,W)$

$p\left(y\right)=\frac{2}{6}\times \frac{8}{12}\times \frac{3}{4}+\frac{2}{6}\times \frac{4}{12}\times \frac{1}{4}$

$\mathrm{p}(y\mid x)=\frac{\frac{2}{6}\times \frac{8}{12}\times \frac{3}{4}+\frac{2}{6}\times \frac{4}{12}\times \frac{1}{4}}{\frac{2}{6}\times \frac{8}{12}\times \frac{3}{4}+\frac{2}{6}\times \frac{4}{12}\times \frac{1}{4}+\frac{4}{6}\times \frac{8}{12}\times \frac{1}{4}}$

$=\frac{0.194444}{0.305555}$

$=0.636365$

Step 3:Final Answer

If one ball is selected from each urn, $0.636365$ is the probability that the ball chosen from urn $A$ was white given that exactly two white balls were selected.