Question

Three persons A, B and C shoot to hit a target. If in trials, A hits the target 4 times in 5 shots, B hits 3 times in 4 shots and C hits 2 times in 3 trials. find the probability that
(a) Exactly two persons hit the target
(b) At least two persons hit the target.

Answer 1

$P\left(A\right)=\frac{4}{5},P\left(B\right)=\frac{3}{4},P\left(C\right)=\frac{2}{3}$
$P\left(\overline{A}\right)=\frac{1}{5},P\left(\overline{B}\right)=\frac{1}{4},P\left(\overline{C}\right)=\frac{1}{3}$
(a) P(Exactly two hit the target)
$P(A\cap B\cap \overline{C})+P(\overline{A}\cap B\cap C)+P(A\cap \overline{B}\cap C)$ [$\because$ A, B, C are independent events]
$=\frac{4}{5}.\frac{3}{4}.\frac{1}{3}+\frac{1}{5}.\frac{3}{4}.\frac{2}{3}+\frac{4}{5}.\frac{1}{4}.\frac{2}{3}$
$=\frac{1}{5}+\frac{1}{10}+\frac{2}{15}=\frac{13}{30}$

(b) P(at least two hit the target)
$=P(A\cap B\cap C)+P(A\cap B\cap \overline{C})+P(\overline{A}\cap B\cap C)+P(A\cap \overline{B}\cap C)$
$=\frac{4}{5}.\frac{3}{4}.\frac{2}{3}+\frac{13}{30}$ [Using (a)]
$=\frac{2}{5}+\frac{13}{30}=\frac{25}{30}=\frac{5}{6}$