Question

In a shooting competition, the probability of a man hitting a target is $\dfrac{3}{5}$. If he fires 4 times, what is the probability of hitting the target.
a) At least twice
b) At most twice

Answer 1

Hint: In this question we have to find the probability of hitting the target at least twice and at most twice. Therefore, we have to first find the events corresponding to the two options and then find out the probability corresponding to each of the options.

Complete step-by-step answer:
It is given that the probability of a man hitting the target is $\dfrac{3}{5}$ and that he fires 4 times. Each time he either can hit the target or miss it. Therefore, the question is of the binomial probability distribution.
In a Binomial probability distribution, in each event (in this case firing by the man) can have two outcomes (in case hitting or missing the target). The formula for getting x successes in n trials (in this case the number of times the man fires) is given by:
$\text{Probability of x successes in n trials=}{}^{n}{{C}_{x}}{{p}^{x}}{{(1-p)}^{n-x}}..................(1.1)$
Where p is the success probability of each event and thus (1-p) is the probability of failure in each event and ${}^{n}{{C}_{x}}=\dfrac{n!}{\left( n-x \right)!x!}$
In this case, it is given that the probability of success (man hitting the target) in each event is $p=\dfrac{3}{5}$
Here the number of trails =the number of times the man fires= n=4
a)The cases corresponding to hitting the target at least twice are hitting the target 2, 3 or 4 times. Therefore,
\[\begin{align}
  & \text{Probability of hitting the target at least twice} \\
 & \text{=Probability of 2 successes+Probability of 3 successes}+\text{Probability of 4 successes} \\
 & \text{=P(2)+P(3)+P(4)=}{}^{4}{{C}_{2}}{{p}^{2}}{{(1-p)}^{4-2}}+{}^{4}{{C}_{3}}{{p}^{3}}{{(1-p)}^{4-3}}+{}^{4}{{C}_{4}}{{p}^{4}}{{(1-p)}^{4-4}} \\
 & =\dfrac{4!}{2!2!}{{\left( \dfrac{3}{5} \right)}^{2}}{{\left( 1-\dfrac{3}{5} \right)}^{2}}+\dfrac{4!}{1!3!}{{\left( \dfrac{3}{5} \right)}^{3}}{{\left( 1-\dfrac{3}{5} \right)}^{1}}+\dfrac{4!}{0!4!}{{\left( \dfrac{3}{5} \right)}^{4}}{{\left( 1-\dfrac{3}{5} \right)}^{0}} \\
 & =0.3456+0.3456+0.1296=0.8208 \\
\end{align}\]
Thus the probability of hitting the target at least twice=0.8208.

b)The cases corresponding to hitting the target at most twice are hitting the target 0, 1 or 2 times. Therefore,
\[\begin{align}
  & \text{Probability of hitting the target at least twice} \\
 & \text{=Probability of 0 successes+Probability of 1 successes}+\text{Probability of 2 successes} \\
 & \text{=P(0)+P(1)+P(2)=}{}^{4}{{C}_{0}}{{p}^{0}}{{(1-p)}^{4-0}}+{}^{4}{{C}_{1}}{{p}^{1}}{{(1-p)}^{4-1}}+{}^{4}{{C}_{2}}{{p}^{2}}{{(1-p)}^{4-2}} \\
 & =\dfrac{4!}{4!0!}{{\left( \dfrac{3}{5} \right)}^{0}}{{\left( 1-\dfrac{3}{5} \right)}^{4}}+\dfrac{4!}{3!1!}{{\left( \dfrac{3}{5} \right)}^{1}}{{\left( 1-\dfrac{3}{5} \right)}^{3}}+\dfrac{4!}{2!2!}{{\left( \dfrac{3}{5} \right)}^{2}}{{\left( 1-\dfrac{3}{5} \right)}^{2}} \\
 & =0.0256+0.1536+0.3456=0.5248 \\
\end{align}\]
Thus the probability of hitting the target at least twice=0.5248.

Note: In this case it is important to remember that the event of an object occurring at least k times means it should occur k or more times whereas that the event of an object occurring at most k times means it should occur k or less number of times.