Question

The probability of a man hitting a target is $\frac{1}{10}$. The least number of shots required, so that the probability of his hitting the target at least once is greater than $\frac{1}{4}$ is

• A. $4$
• B. $3$
• C. $2$
• D. $1$

Answer 1

The correct option is B

$3$

Explanation of the correct option.

Solve the given expression.

Given expression,

Whenever we aim for a target ,there are two possible event to occur ,

either hitting the target which is denoted by $H$

or missing the target which is denoted by $M$

Here, the probability of hitting a target $P\left(H\right)=\frac{1}{10}$

and the probability of missing a target $P\left(M\right)=\frac{9}{10}$

Now ,we have $P\left(H\right)+P\left(M\right)=1$

So, $1-$$\left(probabilityofallshotsresultinfailure\right)>\frac{1}{4}$

$$\begin{gathered} \Rightarrow 1-P{\left(M\right)}^n>\frac{1}{4} \\ \Rightarrow 1-{\left(\frac{9}{10}\right)}^n>\frac{1}{4} \\ \Rightarrow {\left(\frac{9}{10}\right)}^n<\frac{3}{4} \\ \Rightarrow n\ge 3 \end{gathered}$$

Therefore,least number of shots required so that the probability of his hitting the target at least once is greater than $\frac{1}{4}$ is $3$.

Hence option (B) is the correct option.