Bag B_1 contains 6 white and 4 blue balls, Bag B_2 contains 4 white and 6 blue balls, and …

Question

Bag $B_1$ contains 6 white and 4 blue balls, Bag $B_2$ contains 4 white and 6 blue balls, and Bag $B_3$ contains 5 white and 5 blue balls. One of the bags is selected at random and a ball is drawn from it. If the ball is white, then the pro...

Accepted Answer

${E}_{1}:Bag{B}_{1}$ is selected

$\begin{gathered} {B}_{1} \quad  {B}_{2} \quad  {B}_{3} \ 6W4B \quad  4W6B \quad  5W5B \end{gathered}$

${E}_{2}:$ bag ${B}_{2}$ is selected

${E}_{3}:Bag{B}_{3}$ is selected

A : Drawn ball is white

We have to find $P(\frac{{E}_{2}}{A})$

$\begin{gathered} P(\frac{{E}_{2}}{A})=\frac{P({E}_{2})P(\frac{A}{{E}_{2}})}{P({E}_{1})P(\frac{A}{{E}_{1}})+P({E}_{2})P(\frac{A}{{E}_{2}})+P({E}_{3})P(\frac{A}{{E}_{3}})} \ =\frac{\frac{1}{3}	imes \frac{4}{10}}{\frac{1}{3}	imes \frac{6}{10}+\frac{1}{3}	imes \frac{4}{10}+\frac{1}{3}	imes \frac{5}{10}} \ =\frac{4}{15} \end{gathered}$

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