Question

The probability that a man can hit a target is $\dfrac{3}{4}$ . He tries $5$ times. The probability that he will hit the target at least three times is
1. $\dfrac{291}{364}$
2. $\dfrac{371}{464}$
3. $\dfrac{471}{502}$
4. $\dfrac{459}{512}$

Answer 1

Hint: In this problem we need to calculate the probability of hitting the target more than $3$ times. In the problem they have mentioned the probability of hitting as $\dfrac{3}{4}$ which is equal to $p$ . Now we will calculate the probability of not hitting the target which is equal to $q$ by using the formula $p+q=1$ . After having the values of $p$ and $q$, the probability of hitting target at least three times is given by $P\left( X\ge 3 \right)$ is calculated by using the binomial distribution formula $P\left( X=x \right)={}^{n}{{C}_{x}}{{\left( p \right)}^{x}}{{\left( q \right)}^{n-x}}$ .

Complete step by step answer:
Given, the probability of hitting the target is $\dfrac{3}{4}$.
So the value of $p$ will be $p=\dfrac{3}{4}$ .
Now the probability of not hitting the target $q$ is calculated from the formula $p+q=1$, then we will have
$\begin{align}
  & \dfrac{3}{4}+q=1 \\
 & \Rightarrow q=1-\dfrac{3}{4} \\
 & \Rightarrow q=\dfrac{1}{4} \\
\end{align}$
In the problem they have mentioned that the man tries $5$ times. So the value of $n$ will be $n=5$ .
Now the probability of hitting the target at least three times is given by $P\left( X\ge 3 \right)$, then we will have
$P\left( X\ge 3 \right)=P\left( X=3 \right)+P\left( X=4 \right)+P\left( X=5 \right)$
From the binomial distribution we can write $P\left( X=x \right)={}^{n}{{C}_{x}}{{\left( p \right)}^{x}}{{\left( q \right)}^{n-x}}$and use the values we have $n=5$, $p=\dfrac{3}{4}$ and $q=\dfrac{1}{4}$ .
$\begin{align}
  & P\left( X\ge 3 \right)={}^{5}{{C}_{3}}{{\left( \dfrac{3}{4} \right)}^{3}}{{\left( \dfrac{1}{4} \right)}^{2}}+{}^{5}{{C}_{4}}{{\left( \dfrac{3}{4} \right)}^{4}}{{\left( \dfrac{1}{4} \right)}^{1}}+{}^{5}{{C}_{5}}{{\left( \dfrac{3}{4} \right)}^{5}}{{\left( \dfrac{1}{4} \right)}^{0}} \\
 & \Rightarrow P\left( X\ge 3 \right)=\dfrac{270+405+243}{1024} \\
 & \Rightarrow P\left( X\ge 3 \right)=\dfrac{459}{512} \\
\end{align}$
So, the probability of hitting a target at least three times when a man tries $5$ times with a probability of hitting the target $\dfrac{3}{4}$, is $\dfrac{459}{512}$ .
So, the correct answer is “Option 4”.

Note: In this problem they have only asked to calculate the probability of hitting the target at least three times. So we have calculated the value $P\left( X\ge 3 \right)$. If they are asked to calculate the probability of hitting the target at most three times, then we need to calculate the value of $P\left( X\le 3 \right)$ by following the above mentioned procedure.