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 R is:
$a.\dfrac{15}{64}$
$b.\dfrac{39}{64}$
$c.\dfrac{21}{64}$
$d.\dfrac{9}{64}$

Answer 1

We will solve this problem by the multiplication theorem of probability. So, here we have three independent events and let us take probability for hitting target as $P\left( {{P}_{H}} \right)=\dfrac{3}{4},P\left( {{Q}_{H}} \right)=\dfrac{1}{2},P\left( {{R}_{H}} \right)=\dfrac{5}{8}$ . We will also require the concept that the total probability is 1. So, probability of not hitting target for first event will be $P\left( {{\overline{P}}_{H}} \right)=1-\dfrac{3}{4}=\dfrac{1}{4}$ . Similarly, find for other two events too. We have to find the probability that that target is hit by P or Q but not R. So, we will use the formula as $P\left( {{P}_{H}}{{\overline{Q}}_{H}}{{\overline{R}}_{H}} \right)+P\left( {{\overline{P}}_{H}}{{Q}_{H}}{{\overline{R}}_{H}} \right)+P\left( {{P}_{H}}{{Q}_{H}}{{\overline{R}}_{H}} \right)$ and find the answer.

Complete step-by-step answer:
Let us first understand the multiplication theorem of probability: If X and Y are two independent events, then the probability of both will be equal to the product of their individual probabilities.
$P\left( X\cap Y \right)=P\left( X \right)\times P\left( Y \right)$
Here we have three events as
Hitting target by P – ${{P}_{H}}$
Hitting target by Q - ${{Q}_{H}}$
Hitting target by R - ${{R}_{H}}$
Now, the probability of hitting target by P, Q and R are given to us, so we can write it as
$\begin{aligned} & P\left( {{P}_{H}} \right)=\dfrac{3}{4} \\\ & P\left( {{Q}_{H}} \right)=\dfrac{1}{2} \\\ & P\left( {{R}_{H}} \right)=\dfrac{5}{8} \\\ \end{aligned}$
Now, we know that total probability = 1. So, here we will have the probability of hitting target + probability of not hitting the target = 1.
Therefore, we can find the probability of not hitting the target as

& P\left( {{\overline{P}}_{H}} \right)=1-\dfrac{3}{4}=\dfrac{1}{4} \\\ & P\left( {{\overline{Q}}_{H}} \right)=1-\dfrac{1}{2}=\dfrac{1}{2} \\\ & P\left( {{\overline{R}}_{H}} \right)=1-\dfrac{5}{8}=\dfrac{3}{8} \\\ \end{aligned}$$ Now, we will apply the multiplication theorem of probability. According to the question, we want the probability that either P or Q may hit the target but not R. So, we can have cases as P hits the target, but not Q and R. Similarly, Q hits the target, but not P and R. The last case would be both P and Q hit the target but not R. That means we have to find, $P\left( {{P}_{H}}{{\overline{Q}}_{H}}{{\overline{R}}_{H}} \right)+P\left( {{\overline{P}}_{H}}{{Q}_{H}}{{\overline{R}}_{H}} \right)+P\left( {{P}_{H}}{{Q}_{H}}{{\overline{R}}_{H}} \right)$ Now, substituting all the value in the above formula, we get $\begin{aligned} & =\left( \dfrac{3}{4}\times \dfrac{1}{2}\times \dfrac{3}{8} \right)+\left( \dfrac{1}{4}\times \dfrac{1}{2}\times \dfrac{3}{8} \right)+\left( \dfrac{3}{8}\times \dfrac{1}{2}\times \dfrac{3}{8} \right) \\\ & =\dfrac{9}{64}+\dfrac{3}{64}+\dfrac{9}{64} \\\ & =\dfrac{21}{64} \\\ \end{aligned}$ Hence the solution is $\dfrac{21}{64}$. **So, the correct answer is “Option c”.** **Note:** It is necessary that we have to consider all the three cases here and add them together to get the final answer. The common mistake committed by students is by not considering all the cases and just finding $P\left( {{P}_{H}}{{Q}_{H}}{{\overline{R}}_{H}} \right)$ . This will give us only $\dfrac{9}{64}$ as the answer and the student might choose option d as the correct answer. This would have been right if the question was to find the probability of P “and” Q hitting the target and not R. So, this mistake must not be committed in the exams as the student might lose marks.