Question

A can hit a target $4$ times in $5$ shots. B $3$ times in $4$ shots and C twice in $3$ shots. They fire a volley. What is the probability that at least two shots hit the target?

• A. $\frac{1}{6}$
• B. $\frac{4}{6}$
• C. $\frac{5}{6}$
• D. None of these

Answer 1

Answer: (C)

$\frac{5}{6}$

Fire a volley means that $A,BandC$ all try to hit the target simultaneously.

At least two shots hit the target in one of the following ways:

$\left(1\right)$ $A$ and $B$ hit and $C$ fails to hit.

$\left(2\right)$ $A$ and $C$ hit and $B$ fails to hit.

$\left(3\right)$ $B$ and $C$ hit and $A$ fails to hit.

$\left(4\right)$ $A,B,C$ all three hit the target.

The chance of hitting by $A,P\left(A\right)=\frac{4}{5}$ and of not hitting by him $P\left(A^{\prime }\right)=1-\frac{4}{5}=\frac{1}{5}$.

The chance of hitting by $B,P\left(B\right)=\frac{3}{4}$ and of not hitting by him $P\left(B^{\prime }\right)=1-\frac{3}{4}=\frac{1}{4}$.

The chance of hitting by $C,P\left(C\right)=\frac{2}{3}$ and of not hitting by him $P\left(C^{\prime }\right)=1-\frac{2}{3}=\frac{1}{3}$.

The probability of $\left(1\right)$ is $P\left(AB{C}^{\prime }\right)=\frac{4}{5}\times \frac{3}{4}\times \frac{1}{3}=\frac{1}{5}$.

The probability of $\left(2\right)$ is $P\left(AC{B}^{\prime }\right)=\frac{4}{5}\times \frac{2}{3}\times \frac{1}{4}=\frac{2}{15}$.

The probability of $\left(3\right)$ is $P\left(BC{A}^{\prime }\right)=\frac{3}{4}\times \frac{2}{3}\times \frac{1}{5}=\frac{1}{10}$.

The probability of $\left(4\right)$ is $P\left(ABC\right)=\frac{4}{5}\times \frac{3}{4}\times \frac{2}{3}=\frac{2}{5}$

Thus the probability that atleast two shots hit the target is $\frac{1}{5}+\frac{2}{15}+\frac{1}{10}+\frac{2}{5}=\frac{5}{6}$.