Question

The probability that a man can hit a target is $\frac{3}{4}$. He tries $5$times. The probability that he will hit the target at least three times is

• A. $\frac{291}{364}$
• B. $\frac{371}{464}$
• C. $\frac{471}{502}$
• D. $\frac{459}{512}$

Answer 1

The correct option is D

$\frac{459}{512}$

The explanation for the correct answer:

Let $p$ be the probability that the man hits the target

Let $q$ be the probability that the man misses the target$\Rightarrow q=1-p$

Let $n$ be the number of chances

$p=\frac{3}{4},q=\frac{1}{4},n=5$

By using binomial distribution, the required probability

$P\left(X\ge 3\right)=P(X=3)+P(X=4)+P(X=5)$

$$\begin{gathered} ={}^{5}C_3{\left(\frac{3}{4}\right)}^3{\left(\frac{1}{2}\right)}^2+{}^{5}C_4{\left(\frac{3}{4}\right)}^4{\left(\frac{1}{2}\right)}^1+{}^{5}C_5{\left(\frac{3}{4}\right)}^5 \\ =\frac{270+405+243}{1024} \\ =\frac{459}{512} \end{gathered}$$

Hence, the probability of the man hitting the target at least $3$ times is $\frac{459}{512}$

Hence, option(D) is the correct answer