Question

A can hit a target $3$ times in $6$ shots, $B:2$ times in $6$ shots and $C:4$ times in $4$ shots. They fix a volley. What is the probability that at least $2$ shots hit?

Answer 1

Probability of $A=P\left(A\right)=\frac{3}{6}=\frac{1}{2}$

Probability of $B=P\left(B\right)=\frac{2}{6}=\frac{1}{3}$

Probability of $C=P\left(C\right)=\frac{4}{4}=1$

Required probability

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

$=P(A\cap B\cap \overline{C})+P(A\cap \overline{B}\cap C)+P(\overline{A}\cap B\cap C)+P(A\cap B\cap C)]$

$=P\left(A\right)P\left(B\right)P\left(\overline{C}\right)+P\left(A\right)P\left(\overline{B}\right)P\left(C\right)+P\left(\overline{A}\right)P\left(B\right)P\left(C\right)+P\left(A\right)P\left(B\right)P\left(C\right)$ [$A,B,C$ are independents events]

$=\frac{1}{2}\times \frac{1}{3}\times 0+\frac{1}{2}\times \frac{2}{3}\times 1+\frac{1}{2}\times \frac{1}{3}\times 1+\frac{1}{2}\times \frac{1}{3}\times 1$

$=\frac{1}{3}+\frac{1}{6}+\frac{1}{6}=\frac{2}{3}$