Question

Three persons A, B and C shoot to hit the target. If in trials, A hits the target 4 times in 5 shots, B hits 3 times in 4 shots and C hits 2 times in 3 shots. Find the probability that
(a) Exactly two persons hit the target.
(b) At least two people hit the target.

Answer 1

We start solving the problem by finding the probabilities that A, B and C can hit the target. We can see that hitting the target by an individual person is independent and recall the definition of probability of intersection of two independent events. We use this and make calculations to get the probabilities that exactly two persons hit the target and at least two persons hit the target.

Complete step-by-step answer:
According to the problem, we have a three persons A, B and C who are shooting to hit the target. As per the trials, A hits the target 4 times in 5 shots, B hits 3 times in 4 shots and C hits 2 times in 3 shots.
Let us find the probabilities that A, B and C can hit the target using the given values and assume them as $P\left( A \right)$, $P\left( B \right)$ and $P\left( C \right)$.
It is given that A hits the targets 4 times in 5 shoots. So, this gives us the probability that hits the target as $\dfrac{4}{5}$.
∴ $P\left( A \right)=\dfrac{4}{5}$.
It is given that B hits the targets 3 times in 4 shoots. So, this gives us the probability that hits the target as $\dfrac{3}{4}$.
∴ $P\left( B \right)=\dfrac{3}{4}$.
It is given that C hits the targets 2 times in 3 shoots. So, this gives us the probability that hits the target as $\dfrac{2}{3}$.
∴ $P\left( C \right)=\dfrac{2}{3}$.
(a) Now, let us find the probability that exactly two persons hits the target. Let us assume this as $P\left( {{E}_{2}} \right)$.
So, we get $P\left( {{E}_{2}} \right)=P\left( A\cap B\cap {{C}^{c}} \right)+P\left( A\cap {{B}^{c}}\cap C \right)+P\left( {{A}^{c}}\cap B\cap C \right)$.
Here shooting the target by the individual person is independent to each other.
We know that if a, b are two independent events, then $P\left( a\cap b \right)=P\left( a \right).P\left( b \right)$.
$\Rightarrow P\left( {{E}_{2}} \right)=\left( P\left( A \right).P\left( B \right).P\left( {{C}^{c}} \right) \right)+\left( P\left( A \right).P\left( {{B}^{c}} \right).P\left( C \right) \right)+\left( P\left( {{A}^{c}} \right).P\left( B \right).P\left( C \right) \right)$.
We know that $P\left( {{a}^{c}} \right)=1-P\left( a \right)$.
$\Rightarrow P\left( {{E}_{2}} \right)=\left( P\left( A \right).P\left( B \right).\left( 1-P\left( C \right) \right) \right)+\left( P\left( A \right).\left( 1-P\left( B \right) \right).P\left( C \right) \right)+\left( \left( 1-P\left( A \right) \right).P\left( B \right).P\left( C \right) \right)$.
$\Rightarrow P\left( {{E}_{2}} \right)=\left( \dfrac{4}{5}\times \dfrac{3}{4}\times \left( 1-\dfrac{2}{3} \right) \right)+\left( \dfrac{4}{5}\times \left( 1-\dfrac{3}{4} \right)\times \dfrac{2}{3} \right)+\left( \left( 1-\dfrac{4}{5} \right)\times \dfrac{3}{4}\times \dfrac{2}{3} \right)$.
$\Rightarrow P\left( {{E}_{2}} \right)=\left( \dfrac{4}{5}\times \dfrac{3}{4}\times \dfrac{1}{3} \right)+\left( \dfrac{4}{5}\times \dfrac{1}{4}\times \dfrac{2}{3} \right)+\left( \dfrac{1}{5}\times \dfrac{3}{4}\times \dfrac{2}{3} \right)$.
$\Rightarrow P\left( {{E}_{2}} \right)=\left( \dfrac{12}{60} \right)+\left( \dfrac{8}{60} \right)+\left( \dfrac{6}{60} \right)$.
$\Rightarrow P\left( {{E}_{2}} \right)=\dfrac{26}{60}$.
$\Rightarrow P\left( {{E}_{2}} \right)=\dfrac{13}{30}$.
∴ The probability that exactly two hit the target is $\dfrac{13}{30}$.
(b) Now, let us find the probability that at least two persons hits the target. Let us assume this as $P\left( {{A}_{2}} \right)$. This means that we need to find the probability that 2 persons or more persons hit the target.
So, we get $P\left( {{A}_{2}} \right)=P\left( A\cap B\cap {{C}^{c}} \right)+P\left( A\cap {{B}^{c}}\cap C \right)+P\left( {{A}^{c}}\cap B\cap C \right)+P\left( A\cap B\cap C \right)$.
Here shooting the target by the individual person is independent to each other.
We know that if a, b are two independent events, then $P\left( a\cap b \right)=P\left( a \right).P\left( b \right)$.
$\Rightarrow P\left( {{A}_{2}} \right)=\left( P\left( A \right).P\left( B \right).P\left( {{C}^{c}} \right) \right)+\left( P\left( A \right).P\left( {{B}^{c}} \right).P\left( C \right) \right)+\left( P\left( {{A}^{c}} \right).P\left( B \right).P\left( C \right) \right)+\left( P\left( A \right).P\left( B \right).P\left( C \right) \right)$.
We know that $P\left( {{a}^{c}} \right)=1-P\left( a \right)$.
$\Rightarrow P\left( {{A}_{2}} \right)=\left( P\left( A \right).P\left( B \right).\left( 1-P\left( C \right) \right) \right)+\left( P\left( A \right).\left( 1-P\left( B \right) \right).P\left( C \right) \right)+\left( \left( 1-P\left( A \right) \right).P\left( B \right).P\left( C \right) \right)+\left( P\left( A \right).P\left( B \right).P\left( C \right) \right)$.
$\Rightarrow P\left( {{A}_{2}} \right)=\left( \dfrac{4}{5}\times \dfrac{3}{4}\times \left( 1-\dfrac{2}{3} \right) \right)+\left( \dfrac{4}{5}\times \left( 1-\dfrac{3}{4} \right)\times \dfrac{2}{3} \right)+\left( \left( 1-\dfrac{4}{5} \right)\times \dfrac{3}{4}\times \dfrac{2}{3} \right)+\left( \dfrac{4}{5}\times \dfrac{3}{4}\times \dfrac{2}{3} \right)$.
$\Rightarrow P\left( {{A}_{2}} \right)=\left( \dfrac{4}{5}\times \dfrac{3}{4}\times \dfrac{1}{3} \right)+\left( \dfrac{4}{5}\times \dfrac{1}{4}\times \dfrac{2}{3} \right)+\left( \dfrac{1}{5}\times \dfrac{3}{4}\times \dfrac{2}{3} \right)+\left( \dfrac{4}{5}\times \dfrac{3}{4}\times \dfrac{2}{3} \right)$.
$\Rightarrow P\left( {{A}_{2}} \right)=\left( \dfrac{12}{60} \right)+\left( \dfrac{8}{60} \right)+\left( \dfrac{6}{60} \right)+\left( \dfrac{24}{60} \right)$.
$\Rightarrow P\left( {{A}_{2}} \right)=\dfrac{50}{60}$.
$\Rightarrow P\left( {{A}_{2}} \right)=\dfrac{5}{6}$.
∴ The probability that at least two hit the target is $\dfrac{5}{6}$.

Note: We can also find the probability of at least two hit the target by subtracting the probabilities that utmost one person hit the target from 1. (calculation shown below)
$\Rightarrow P\left( {{A}_{2}} \right)=\left( P\left( {{A}^{c}} \right).P\left( {{B}^{c}} \right).P\left( {{C}^{c}} \right) \right)+\left( P\left( {{A}^{c}} \right).P\left( {{B}^{c}} \right).P\left( C \right) \right)+\left( P\left( A \right).P\left( {{B}^{c}} \right).P\left( {{C}^{c}} \right) \right)+\left( P\left( {{A}^{c}} \right).P\left( B \right).P\left( {{C}^{c}} \right) \right)$.
$\Rightarrow P\left( {{A}_{2}} \right)=1-\left( \left( \left( 1-\dfrac{4}{5} \right)\times \left( 1-\dfrac{3}{4} \right)\times \left( 1-\dfrac{2}{3} \right) \right)+\left( \left( 1-\dfrac{4}{5} \right)\times \left( 1-\dfrac{3}{4} \right)\times \dfrac{2}{3} \right)+\left( \dfrac{4}{5}\times \left( 1-\dfrac{3}{4} \right)\times \left( 1-\dfrac{2}{3} \right) \right)+\left( \left( 1-\dfrac{4}{5} \right)\times \dfrac{3}{4}\times \left( 1-\dfrac{2}{3} \right) \right) \right)$.
$\Rightarrow P\left( {{A}_{2}} \right)=1-\left( \left( \dfrac{1}{5}\times \dfrac{1}{4}\times \dfrac{1}{3} \right)+\left( \dfrac{1}{5}\times \dfrac{1}{4}\times \dfrac{2}{3} \right)+\left( \dfrac{4}{5}\times \dfrac{1}{4}\times \dfrac{1}{3} \right)+\left( \dfrac{1}{5}\times \dfrac{3}{4}\times \dfrac{1}{3} \right) \right)$.
$\Rightarrow P\left( {{A}_{2}} \right)=1-\left( \left( \dfrac{1}{60} \right)+\left( \dfrac{2}{60} \right)+\left( \dfrac{4}{60} \right)+\left( \dfrac{3}{60} \right) \right)$.
$\Rightarrow P\left( {{A}_{2}} \right)=1-\left( \dfrac{1+2+4+3}{60} \right)$.
$\Rightarrow P\left( {{A}_{2}} \right)=1-\left( \dfrac{10}{60} \right)$.
$\Rightarrow P\left( {{A}_{2}} \right)=1-\dfrac{1}{6}$.
$\Rightarrow P\left( {{A}_{2}} \right)=\dfrac{6-1}{6}$.
$\Rightarrow P\left( {{A}_{2}} \right)=\dfrac{5}{6}$.