Question

Three persons P, Q and R independently try to hit a target. If the probabilities of their hitting the target are $\dfrac{3}{4}$, $\dfrac{1}{2}$ and $\dfrac{5}{8}$ respectively, then the probability that the target is hit by P or Q but not by R is$$$$
A. \dfrac{39}{64}$$$$$ B. \dfrac{21}{64} C. $\dfrac{9}{64}
D. $\dfrac{15}{64}$$$$$

Answer 1

Use the concept of compliments for mutually exclusive events in probability to find out the probability of not hitting a target. Then use conditional probability to find out the required result.

Complete step by step answer:
Mutually exclusive events are events which do not occur at the same time. It means if there are 2 only possible events say $A$ and $\text{B}$ if mutually exclusive then
$P\left( A\bigcap B \right)=0,P\left( A\bigcup B \right)=P\left( A \right)+P\left( B \right)=1$
Let us assign the probability of person P hitting the target as ${{P}_{H}}$ and probability of not hitting the target as ${{P}_{N}}$. Similarly we assign the probability of the person Q hitting the target as ${{Q}_{H}}$ and the probability of not hitting the target as ${{Q}_{N}}$. Then we assign the probability of the person R hitting the target as ${{R}_{H}}$ and not hitting the target as ${{R}_{N}}$.$$$$
A person can either hit the target or not hit the target. So hitting or not hitting the target are two mutually exclusive events.
It is given in the question that ${{P}_{H}}=\dfrac{3}{4},{{Q}_{H}}=\dfrac{1}{2},{{R}_{H}}=\dfrac{5}{8}$. So using the formula for two mutually exclusive events we get,

& {{P}_{H}}+{{P}_{N}}=1\Rightarrow {{P}_{N}}=1-{{P}_{H}}=\dfrac{1}{4} \\\ & {{Q}_{H}}+{{Q}_{N}}=1\Rightarrow {{Q}_{N}}=1-{{Q}_{H}}=\dfrac{1}{2} \\\ & {{R}_{H}}+{{R}_{N}}=1\Rightarrow {{R}_{N}}=1-{{R}_{H}}=\dfrac{3}{8} \\\ \end{aligned}$$ Independent events are events whose occurrence one does not affect the occurrence of other events even if they occur at the same time. Mathematically, $A$ and \$text{B}$ are two independent events then from the theory of conditional probability , the probability of $A$ and $\text{B}$ happening together is given by, $$P\left( A\bigcap B \right)=P\left( A \right).P\left( B \right)$$ Here the hitting or not hitting the target of one person does not change the hitting or not hitting of others. So all these events are independent. We are asked to find out the probability of the target being hit by either by P or Q but not by R. So we get three situations, where probability of (i) only P hitting the target and Q and R not hitting is $${{P}_{H}}{{Q}_{N}}{{R}_{N}}=\dfrac{3}{4}\left( \dfrac{1}{2} \right)\left( \dfrac{3}{8} \right)$$ (ii)both P and Q hitting the target and R not hitting is $${{P}_{H}}{{Q}_{H}}{{R}_{N}}=\dfrac{3}{4}\left( \dfrac{1}{2} \right)\left( \dfrac{3}{8} \right)$$ (iii) only Q hitting the target and P and R not hitting is $${{P}_{N}}{{Q}_{H}}{{R}_{N}}=\dfrac{1}{4}\left( \dfrac{1}{2} \right)\left( \dfrac{3}{8} \right)$$ So total probability is $${{P}_{H}}{{Q}_{N}}{{R}_{N}}+{{P}_{H}}{{Q}_{H}}{{R}_{N}}+{{P}_{N}}{{Q}_{H}}{{R}_{N}}=\dfrac{3}{4}\left( \dfrac{1}{2} \right)\left( \dfrac{3}{8} \right)+\dfrac{3}{4}\left( \dfrac{1}{2} \right)\left( \dfrac{3}{8} \right)+\dfrac{1}{4}\left( \dfrac{1}{2} \right)\left( \dfrac{3}{8} \right)=\dfrac{21}{64}$$ **So, the correct answer is “Option B”.** **Note:** It is to be noted that mutually exclusive events and independent events do not necessarily imply each other. So we should be careful while using their formulas in the problems of probability.