Question

Three persons $A,B$ and $C$ shoot to hit a target. If $A$ hits the target four times in five trials, $B$ hits it three times in four trials and $C$ hits it two times in three trials, find the probability that:
i. Exactly two persons hit the target.
ii. At least two persons hit the target.
iii. None hit the target.

Answer 1

Given, $P\left(A\right)=\frac{4}{5},P\left(B\right)=\frac{3}{4},P\left(C\right)=\frac{2}{3}$
(i) $P($exactly two persons hit the target$)$
$=P(A\cap B\cap \overline{C})+P(A\cap \overline{B}\cap C)+P(\overline{A}\cap B\cap C)$

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

$=\frac{4}{5}.\frac{3}{4}\left(\begin{array}{c}1-\frac{2}{3}\end{array}\right)+\frac{4}{5}\left(\begin{array}{c}1-\frac{3}{4}\end{array}\right).\frac{2}{3}+\left(\begin{array}{c}1-\frac{4}{5}\end{array}\right).\frac{3}{4}.\frac{2}{3}$
$=\frac{4}{5}.\frac{3}{4}.\frac{1}{3}+\frac{4}{5}.\frac{1}{4}.\frac{2}{3}+\frac{1}{5}.\frac{3}{4}.\frac{2}{3}$

$=\frac{12+8+6}{60}=\frac{26}{60}=\frac{13}{30}$.

(ii) $P($at least two persons hit the target$)$
$=P($all three persons hit the target$)+P($exactly two persons hit the target$)$

$=P(A\cap B\cap C)+\frac{13}{30}$ [Using $\left(i\right)$]

$=P\left(A\right)P\left(B\right)P\left(C\right)+\frac{13}{30}$

$=\frac{4}{5}.\frac{3}{4}.\frac{2}{3}+\frac{13}{30}$
Ans. $=\frac{2}{5}+\frac{13}{30}=\frac{12+13}{30}=\frac{25}{30}=\frac{5}{6}$.

(iii) $P($None hit the target$)$
$=P(\overline{A}\cap \overline{B}\cap \overline{C})$

$=P\left(\overline{A}\right).P\left(\overline{B}\right).P\left(\overline{C}\right)$
$=\left(\begin{array}{c}1-\frac{4}{5}\end{array}\right).\left(\begin{array}{c}1-\frac{3}{4}\end{array}\right).\left(\begin{array}{c}1-\frac{2}{3}\end{array}\right)$

Ans. $=\frac{1}{5}.\frac{1}{4}.\frac{1}{3}=\frac{1}{60}$.