Question

In a shooting test the probability of A,B, C hitting the targets are $\dfrac{1}{2}$, $\dfrac{2}{3}$ and $\dfrac{3}{4}$ respectively. If all of them fire at the same target, find the probability that
(i) only one of them hits the target
(ii) at least one of them hits the target

Answer 1

Hint: To calculate the probability that only one of them hits the target, we have to consider the probability that one of them hits the target and the other two miss the target. We have to do this for all of them, that is A, B and C. Then, to find the probability that at least one of them hits the target, we will subtract the probability that none of them hits the target from 1.This will give us the required answer.

Complete step by step answer:
The given probabilities are of independent events. Let the probability that A hits the target be denoted by P(A). Similarly for B and C, we have P(B) and P(C). So we have$\text{P}\left( \text{A} \right)\text{ =}\dfrac{1}{2}$, $\text{P}\left( \text{B} \right)\text{ =}\dfrac{2}{3}$ and $\text{P}\left( \text{C} \right)\text{ =}\dfrac{3}{4}$. Now, we will find the probability that only one of them hits the target.
For this, we have to consider that one of them hits the target and the other two miss the target. The probability of A missing the target is given as,
$\text{P}\left( \overline{\text{A}} \right)\text{=1}-\text{P}\left( \text{A} \right)$
Similarly, the probability of B missing the target we have $\text{P}\left( \overline{\text{B}} \right)\text{=1}-\text{P}\left( \text{B} \right)$ and the probability that C misses the target is $\text{P}\left( \overline{\text{C}} \right)\text{=1}-\text{P}\left( \text{C} \right)$. Now, let the probability that only one of them hits the target be denoted by $\text{P}\left( {{\text{E}}_{1}} \right)$. So, we have
$\text{P}\left( {{\text{E}}_{1}} \right)=\text{P}\left( \text{A}\cap \overline{\text{B}}\cap \overline{\text{C}} \right)+\text{P}\left( \overline{\text{A}}\cap \text{B}\cap \overline{\text{C}} \right)+\text{P}\left( \overline{\text{A}}\cap \overline{\text{B}}\cap \text{C} \right)$
Since the events A, B and C are independent, the probability of their intersections is the same as the probability of their products. This implies that we have the following equation,
\[\text{P}\left( {{\text{E}}_{1}} \right)=\text{P}\left( \text{A} \right)\text{P}\left( \overline{\text{B}} \right)\text{P}\left( \overline{\text{C}} \right)+\text{P}\left( \overline{\text{A}} \right)\text{P}\left( \text{B} \right)\text{P}\left( \overline{\text{C}} \right)+\text{P}\left( \overline{\text{A}} \right)\text{P}\left( \overline{\text{B}} \right)\text{P}\left( \text{C} \right)\]
Now, we will substitute $\text{P(}\overline{\text{A}}\text{) = 1}-\text{P(A)}$, $\text{P(}\overline{\text{B}}\text{) = 1}-\text{P(B)}$ and $\text{P(}\overline{\text{C}}\text{) = 1}-\text{P(C)}$. So, we have
$\text{P}\left( {{\text{E}}_{1}} \right)=\text{P}\left( \text{A} \right)\left( 1-\text{P}\left( \text{B} \right) \right)\left( 1-\text{P}\left( \text{C} \right) \right)+\left( 1-\text{P}\left( \text{A} \right) \right)\text{P}\left( \text{B} \right)\left( 1-\text{P}\left( \text{C} \right) \right)+\left( 1-\text{P}\left( \text{A} \right) \right)\left( 1-\text{P}\left( \text{B} \right) \right)\text{P}\left( \text{C} \right)$
We know that $\text{P}\left( \text{A} \right)\text{ =}\dfrac{1}{2}$, $\text{P}\left( \text{B} \right)\text{ =}\dfrac{2}{3}$ and $\text{P}\left( \text{C} \right)\text{ =}\dfrac{3}{4}$. Substituting these values in the above equation, we get
$\begin{align}
  & \text{P}\left( {{\text{E}}_{1}} \right)=\left( \dfrac{1}{2} \right)\left( 1-\dfrac{2}{3} \right)\left( 1-\dfrac{3}{4} \right)+\left( 1-\dfrac{1}{2} \right)\left( \dfrac{2}{3} \right)\left( 1-\dfrac{3}{4} \right)+\left( 1-\dfrac{1}{2} \right)\left( 1-\dfrac{2}{3} \right)\left( \dfrac{3}{4} \right) \\
 & \Rightarrow \text{P}\left( {{\text{E}}_{1}} \right)=\left( \dfrac{1}{2} \right)\left( \dfrac{1}{3} \right)\left( \dfrac{1}{4} \right)+\left( \dfrac{1}{2} \right)\left( \dfrac{2}{3} \right)\left( \dfrac{1}{4} \right)+\left( \dfrac{1}{2} \right)\left( \dfrac{1}{3} \right)\left( \dfrac{3}{4} \right) \\
 & \Rightarrow \text{P}\left( {{\text{E}}_{1}} \right)=\dfrac{1}{24}+\dfrac{2}{24}+\dfrac{3}{24} \\
 & \Rightarrow \text{P}\left( {{\text{E}}_{1}} \right)=\dfrac{6}{24} \\
 & \therefore \text{P}\left( {{\text{E}}_{1}} \right)=\dfrac{1}{4} \\
\end{align}$
Therefore, the probability that only one of them hits the target is $\dfrac{1}{4}$.
Now, we have to find the probability that atleast one of them hits the target. Let us denote the probability of this event as $\text{P}\left( {{\text{E}}_{2}} \right)$. To calculate this probability, we will subtract the probability that none of them hit the target from 1.
The probability that none of them hit the target is given by $\text{P}\left( \overline{\text{A}}\cap \overline{\text{B}}\cap \overline{\text{C}} \right)$. Therefore, we have
\[\begin{align}
  & \text{P}\left( {{\text{E}}_{2}} \right)=1-\text{P}\left( \overline{\text{A}}\cap \overline{\text{B}}\cap \overline{\text{C}} \right) \\
 & \Rightarrow \text{P}\left( {{\text{E}}_{2}} \right)=1-\text{P}\left( \overline{\text{A}} \right)\text{P}\left( \overline{\text{B}} \right)\text{P}\left( \overline{\text{C}} \right) \\
 & \Rightarrow \text{P}\left( {{\text{E}}_{2}} \right)=1-\left( \text{1}-\text{P}\left( \text{A} \right) \right)\left( \text{1}-\text{P}\left( \text{B} \right) \right)\left( \text{1}-\text{P}\left( \text{C} \right) \right) \\
\end{align}\]
Substituting the given values of the probabilities, we get
\[\begin{align}
  & \text{P}\left( {{\text{E}}_{2}} \right)=1-\left( \text{1}-\dfrac{1}{2} \right)\left( \text{1}-\dfrac{2}{3} \right)\left( \text{1}-\dfrac{3}{4} \right) \\
 & \Rightarrow \text{P}\left( {{\text{E}}_{2}} \right)=1-\left( \dfrac{1}{2} \right)\left( \dfrac{1}{3} \right)\left( \dfrac{1}{4} \right) \\
 & \Rightarrow \text{P}\left( {{\text{E}}_{2}} \right)=1-\dfrac{1}{24} \\
 & \therefore \text{P}\left( {{\text{E}}_{2}} \right)=\dfrac{23}{24} \\
\end{align}\]

Hence, the probability that at least one of them hits the target is $\dfrac{23}{24}$.

Note: In this type of question, it is extremely important that we interpret the events stated correctly. We should be able to realize the relations between the events described so that we can calculate the probabilities of the related events occurring by using the given probabilities. It is also essential to understand the difference between independent events and dependent events.