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 probability, that the ball is drawn from Bag $B_2$, is :

Question

Aicharya · Accepted Answer

$E_1: B a g B_1$ is selected
$\begin{array}{lll}
B_1 & B_2 & B_3 \\ \text { 6W4B } & 4 \mathrm{~W} 6 \mathrm{~B} & 5 \mathrm{~W} 5 \mathrm{~B}
\end{array}$
$\mathrm{E}_2:$ bag $\mathrm{B}_2$ is selected
$E_3: B a g B_3$ is selected
A : Drawn ball is white
We have to find $\mathrm{P}\left(\frac{\mathrm{E}_2}{\mathrm{~A}}\right)$
$P\left(\frac{E_2}{A}\right)=\frac{P\left(E_2\right) P\left(\frac{A}{E_2}\right)}{P\left(E_1\right) P\left(\frac{A}{E_1}\right)+P\left(E_2\right) P\left(\frac{A}{E_2}\right)+P\left(E_3\right) P\left(\frac{A}{E_3}\right)}$
$\begin{aligned} & =\frac{\frac{1}{3} \times \frac{4}{10}}{\frac{1}{3} \times \frac{6}{10}+\frac{1}{3} \times \frac{4}{10}+\frac{1}{3} \times \frac{5}{10}} \\ & =\frac{4}{15}\end{aligned}$ ~

Bag $B_1$ contains 6 white and 4 blue balls, Bag $B_2$ contains 4 white and 6 blue balls, and Bag $B_3$…

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