Bag $A$ contains $5$ white, $6$ black balls and bag $B$ contains $4$ white, $5$ black balls. One bag is selected at random and one ball is drawn from it. Find the probability that it is white.
Let $A_1$ be the event of selecting bag $A$.
Let $A_2$ be the event of selecting bag $B$.
Let $B$ be the event of selecting a white ball.
To find: P(of selecting a white ball) $= P \left(B\right)$
Now, $A_1$ and $A_2$ are mutually exclusive and exhaustive events.
$\therefore \;$ $P \left(B\right) = P \left(A_1\right) \times P \left(B | A_1\right) + P \left(A_2\right) \times P \left(B | A_2\right)$ $\;\;\; \cdots \; (1)$
Probability of selecting bag $A = P \left(A_1\right) = \dfrac{1}{2}$
Probability of selecting bag $B = P \left(A_2\right) = \dfrac{1}{2}$
P(selecting a white ball when bag $A$ is selected) $= P \left(B | A_1\right)$
$P \left(B | A_1\right) = \dfrac{{^{5}}{C}_{1}}{{^{11}}{C}_{1}} = \dfrac{5}{11}$
P(selecting a white ball when bag $B$ is selected) $= P \left(B | A_2\right)$
$P \left(B | A_2\right) = \dfrac{{^{4}}{C}_{1}}{{^{9}}{C}_{1}} = \dfrac{4}{9}$
Substituting the values of $P \left(A_1\right)$, $P \left(A_2\right)$, $P \left(B | A_1\right)$ and $P \left(B | A_2\right)$ in equation $(1)$ we have,
$P \left(B\right) = \left(\dfrac{1}{2} \times \dfrac{5}{11}\right) + \left(\dfrac{1}{2} \times \dfrac{4}{9}\right) = \dfrac{89}{198}$