Question

A man makes attempts to hit the target. The probability of hitting the target is $\dfrac{3}{5}$. Then, the probability that A hits the target exactly $2$ times in $5$ attempts is:
(A) $\dfrac{{144}}{{625}}$
(B) $\dfrac{{72}}{{3125}}$
(C) $\dfrac{{216}}{{625}}$
(D) None of these

Answer 1

In the given question, we are provided with the probability with which a man attempts to hit a target and we have to find the probability that he hits exactly two out of five times. So, we will make use of the multiplication rule of probability to find the probability of two events happening one after another. We will first find the number of ways in which two attempts can be chosen out of a total five attempts and then find the required probability.

Complete step-by-step solution:
So, we have the probability of hitting a target as $\dfrac{3}{5}$.
Then, we have to find the probability that the target is hit exactly two times in five attempts. So, this means that the man hits the target successfully two times and fails to hit the target three times.
We know that the sum of probabilities of all the possibilities of an event is one.
So, the probability of not hitting the target $= 1 - \dfrac{3}{5} = \dfrac{2}{5}$
Now, we know that we can choose two out of five attempts in $^5{C_2}$ ways.
Now, we know that the probability of success is $\dfrac{3}{5}$ and probability of failure is $\dfrac{2}{5}$.
Now, we use the binomial probability distribution with parameters n, r, p and q representing the total number of trials, number of successful trials, probability of success and probability of failure respectively. The formula for binomial probability distribution is $^n{C_r}{p^r}{q^{\left( {n - r} \right)}}$.
So, we get the required probability of hitting the target exactly $2$ times in $5$ attempts as $\left( {^5{C_2}} \right) \times {\left( {\dfrac{3}{5}} \right)^2} \times {\left( {\dfrac{2}{5}} \right)^3}$ .
So, we get, required probability $= \left( {^5{C_2}} \right) \times {\left( {\dfrac{3}{5}} \right)^2} \times {\left( {\dfrac{2}{5}} \right)^3}$
We know the combination formula $^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}$. So, we get,
$= \left( {\dfrac{{5!}}{{2! \times 3!}}} \right) \times \left( {\dfrac{9}{{25}}} \right) \times \left( {\dfrac{8}{{125}}} \right)$
We know values of the factorials of $5$, $2$ and $3$.
$= \left( {\dfrac{{120}}{{2 \times 6}}} \right) \times \left( {\dfrac{9}{{25}}} \right) \times \left( {\dfrac{8}{{125}}} \right)$
Cancelling the common factors in numerator and denominator, we get,
$= 2 \times \left( {\dfrac{9}{5}} \right) \times \left( {\dfrac{8}{{125}}} \right)$
Simplifying the calculations,
$= \dfrac{{144}}{{625}}$
Hence, the probability that A hits the target exactly $2$ times in $5$ attempts is $\dfrac{{144}}{{625}}$.
Therefore, option (A) is the correct answer.

Note: We must know the multiplication rule of probability for finding probability of the events happening one after another. One should know the applications of simplification rules to simplify the calculations. We should remember the combination formula $^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}$ to evaluate the number of ways of choosing two out of five attempts.