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

Accepted Answer

$E_1: B a g B_1$ is selected

$$\[\begin{array}{lll} B_1 & B_2 & B_3 \ 6 \mathrm{~W} 4 \mathrm{~B} & 4 \mathrm{~W} 6 \mathrm{~B} & 5 \mathrm{~W} 5 \mathrm{~B} \end{array}$$\]

$E_2:$ bag $B_2$ is selected

$E_3: B a g B_3$ is selected

A : Drawn ball is white

We have to find $P\left(\frac{E_2}{A}\right)$

$$\begin{aligned}
& P\left(\frac{E_2}{A}\right)=\frac{P\left(E_2\right) P\left(\frac{A}{E_2}\right)}{P\left(E_1\right) P\left(\frac{A}{E_1}\right)+P\left(E_2\right) P\left(\frac{A}{E_2}\right)+P\left(E_3\right) P\left(\frac{A}{E_3}\right)} \
& =\frac{\frac{1}{3} \times \frac{4}{10}}{\frac{1}{3} \times \frac{6}{10}+\frac{1}{3} \times \frac{4}{10}+\frac{1}{3} \times \frac{5}{10}} \
& =\frac{4}{15}
\end{aligned}$$

Solutions · 1