Question

The probability that $A$ hits a target is $\frac{1}{3}$ and the probability that $B$ hits it is $\frac{2}{5}$ . What is the probability that the target will be hit, if each one of $A$ and $B$ shoots the target ?

• A. $\frac{5}{6}$
• B. $\frac{3}{5}$
• C. $\frac{11}{15}$
• D. $\frac{1}{6}$

Answer 1

The correct option is B

$\frac{3}{5}$

Explanation of the correct option:

Finding the required probability:

Let $E$ be the event that the target is hit by $A$, and $F$ be the event that the target is hit by $B$.

$\Rightarrow P\left(E\right)=\frac{1}{3}andP\left(F\right)=\frac{2}{5}$

The probability that the target is not hit by $A$ , $P\left(\overline{E}\right)=1-\frac{1}{3}=\frac{2}{3}\left[\mathrm{by}P\left(A\right)+P\left(\overline{A}\right)=1\right]$.

Similarly, the probability that the target is not hit by $B$, $P\left(\overline{F}\right)=1-\frac{2}{5}=\frac{3}{5}$

We have to find the probability that the target will be hit. This could be possible if $A$ hits the target and $B$ does not hit the target or $B$ hits the target and $A$ does not hit the target or both $A$ and $B$ hit the target.

Therefore, the required probability will be $=P\left(E\right)\times P\left(\overline{F}\right)+P\left(F\right)\times P\left(\overline{E}\right)+P\left(E\right)\times P\left(F\right)$

$\begin{array}{rcl}P\left(E\right)\times P\left(\overline{F}\right)+P\left(F\right)\times P\left(\overline{E}\right)+P\left(E\right)\times P\left(F\right)& =& \left[\frac{1}{3}\times \frac{3}{5}\right]+\left[\frac{2}{5}\times \frac{2}{3}\right]+\left[\frac{1}{3}\times \frac{2}{5}\right]\\ & =& \frac{3}{15}+\frac{4}{15}+\frac{2}{15}\\ & =& \frac{9}{15}\\ & =& \frac{3}{5}\end{array}$

Hence, the correct answer is option B.