Bag $B_1$ contains 4 white and 2 black balls. Bag $B_2$ contains 3 white and 4 black balls. A bag is chosen…
Bag $B_1$ contains 4 white and 2 black balls. Bag $B_2$ contains 3 white and 4 black balls. A bag is chosen at random and a ball is drawn from it at random, then the probability that the ball drawn is white, is
$\frac{1}{42}$
$\frac{42}{32}$
$\frac{33}{42}$
$\frac{23}{42}$
Solution
Let $P\left(B_1\right), P\left(B_2\right)$ and $P(W)$ are the probabilty of selecting bag 1 , bag 2 and white bag respectively.
$
P\left(B_1\right)=\frac{1}{2}=P\left(B_2\right)
$
$
\begin{aligned}
& P\left(\frac{W}{B_1}\right)=\frac{4}{6}=\frac{2}{3} \\
& P\left(\frac{W}{B_2}\right)=\frac{3}{7}
\end{aligned}
$
Now, from the total probability theorem:
$
\begin{aligned}
& P(W)=P\left(\frac{W}{B_1}\right) P\left(B_1\right)+\mathrm{P}\left(\frac{W}{B_2}\right) P\left(B_2\right) \\
& =\frac{1}{2} \times \frac{2}{3}+\frac{1}{2} \times \frac{3}{7}=\frac{23}{42}
\end{aligned}
$