Probability

A can hit a target $4$ times in $5$ shots, B $3$ times in $4$ shots and C $2$ times in $3$ shots. They fire a volley. What is the chance that the target is damaged by exactly $2$ hits?


Let $A$ be the event that A hits the target.

Let $B$ be the event that B hits the target.

Let $C$ be the event that C hits the target.

Then, $P \left(A\right) = \dfrac{4}{5}$, $\;$ $P \left(\overline{A}\right) = 1 - P \left(A\right) = 1 - \dfrac{4}{5} = \dfrac{1}{5}$

$P \left(B\right) = \dfrac{3}{4}$, $\;$ $P \left(\overline{B}\right) = 1 - P \left(B\right) = 1 - \dfrac{3}{4} = \dfrac{1}{4}$

$P \left(C\right) = \dfrac{2}{3}$, $\;$ $P \left(\overline{C}\right) = 1 - P \left(C\right) = 1 - \dfrac{2}{3} = \dfrac{1}{3}$

Now, probability that the target is damaged by exactly $2$ hits

$= P \left[\left(A \cap B \cap \overline{C}\right) \cup \left(A \cap \overline{B} \cap C\right) \cup \left(\overline{A} \cap B \cap C\right)\right]$

$= P \left(A\right) \times P \left(B\right) \times P \left(\overline{C}\right) + P \left(A\right) \times P \left(\overline{B}\right) \times P \left(C\right) + P \left(\overline{A}\right) \times P \left(B\right) \times P \left(C\right)$

[$\because \;$ $A$, $B$, $C$ are independent events, $\overline{A}$, $\overline{B}$, $\overline{C}$ are also independent events]

$= \left(\dfrac{4}{5} \times \dfrac{3}{4} \times \dfrac{1}{3}\right) + \left(\dfrac{4}{5} \times \dfrac{1}{4} \times \dfrac{2}{3}\right) + \left(\dfrac{1}{5} \times \dfrac{3}{4} \times \dfrac{2}{3}\right) $

$= \dfrac{26}{60}$

$= \dfrac{13}{30}$