Question

Three shots are fired at a target in succession. The probabilities of a hit in the first shot is ${\displaystyle \frac{1}{2}}$, in the second ${\displaystyle \frac{2}{3}}$ and in the third shot is ${\displaystyle \frac{3}{4}}$, in case of exactly one hit, the probability of destroying the target is ${\displaystyle \frac{1}{3}}$ and in the case of exactly two hits ${\displaystyle \frac{7}{11}}$ and in the case of three hits is $1.0$. Find the probability of destroying the target in three shots
A.${\displaystyle \frac{3}{4}}$
B.${\displaystyle \frac{3}{8}}$
C.${\displaystyle \frac{5}{8}}$
D.${\displaystyle \frac{4}{5}}$

Answer 1

Hint: Here, we will first find the probability of destroying the target in one shot, two shots and three shots respectively. We will then add them together to find the required probability of destroying the target in three shots. Probability is defined as the certainty of occurrence of an event.

Complete step-by-step answer:
According to the question,
Three shots are fired at a target in succession
Let $A$ represents that the target was hit in the first shot. Therefore,
$A={\displaystyle \frac{1}{2}}$
Let $B$ represents that the target was hit in the second shot. Therefore,
$B={\displaystyle \frac{2}{3}}$
Let $C$ represents that the target was hit in the third shot. Therefore,
$C={\displaystyle \frac{3}{4}}$
Now, let $E_1$ be the event that the target was destroyed in exactly 1 hit. Therefore,
$E_1={\displaystyle \frac{1}{3}}$
Let $E_2$ be the event that the target was destroyed in exactly 2 hits. Therefore,
 $E_2={\displaystyle \frac{7}{11}}$
And, let $E_3$ be the event that the target was destroyed in exactly 3 hits. Therefore,
 $E_3=1$
Now, in order to find the probability of destroying the target in three shots, we will find the probability of destroying the target in exactly one shot.
Now, it is possible that we destroy it in the first or second or on the third shot.
Thus, this can be written as:
$P\left(E_1\right)=P\left(E_{1}A\overline{BC}\cup E_1\overline{AB}C\cup E_1\bar{A}B\bar{C}\right)$
This shows the three cases where the target was destroyed in exactly one shot.
Substituting the given values in the above equation, we get,
$\Rightarrow P\left(E_1\right)={\displaystyle \frac{1}{3}}\left[{\displaystyle \frac{1}{2}}\times {\displaystyle \frac{1}{3}}\times {\displaystyle \frac{1}{4}}+{\displaystyle \frac{1}{2}}\times {\displaystyle \frac{1}{3}}\times {\displaystyle \frac{3}{4}}+{\displaystyle \frac{1}{2}}\times {\displaystyle \frac{2}{3}}\times {\displaystyle \frac{1}{4}}\right]$
Multiplying the terms, we get
$\Rightarrow P\left(E_1\right)={\displaystyle \frac{1}{3}}\left[{\displaystyle \frac{1}{24}}+{\displaystyle \frac{3}{24}}+{\displaystyle \frac{2}{24}}\right]$
Adding the terms inside the bracket, we get
$\Rightarrow P\left(E_1\right)={\displaystyle \frac{1}{3}}\left[{\displaystyle \frac{6}{24}}\right]={\displaystyle \frac{1}{12}}$
Similarly, we will find the probability of destroying the target in exactly two shots.
Now, it is possible that we destroy it in the first and second or second and third or on the third and first shots.
Thus, this can be written as:
$P\left(E_2\right)=P\left(E_2\bar{A}BC\cup E_{2}AB\bar{C}\cup E_{2}A\bar{B}C\right)$
Substituting the given values in the above equation, we get,
$\Rightarrow P\left(E_2\right)={\displaystyle \frac{7}{11}}\left[{\displaystyle \frac{1}{2}}\times {\displaystyle \frac{2}{3}}\times {\displaystyle \frac{3}{4}}+{\displaystyle \frac{1}{2}}\times {\displaystyle \frac{2}{3}}\times {\displaystyle \frac{1}{4}}+{\displaystyle \frac{1}{2}}\times {\displaystyle \frac{1}{3}}\times {\displaystyle \frac{3}{4}}\right]$
Multiplying the terms, we get
$\Rightarrow P\left(E_2\right)={\displaystyle \frac{7}{11}}\left[{\displaystyle \frac{6}{24}}+{\displaystyle \frac{2}{24}}+{\displaystyle \frac{3}{24}}\right]$
Adding the terms inside the bracket, we get
$\Rightarrow P\left(E_2\right)={\displaystyle \frac{7}{11}}\left[{\displaystyle \frac{11}{24}}\right]={\displaystyle \frac{7}{24}}$
Now, we will find the probability of destroying the target in exactly three shots.
It is possible that we destroy it in all the three shots. Hence, only one case is possible.
Thus, this can be written as:
$P\left(E_3\right)=P\left(E_{3}ABC\right)$
Substituting the given values in the above equation, we get
$\Rightarrow P\left(E_3\right)=1\left[{\displaystyle \frac{1}{2}}\times {\displaystyle \frac{2}{3}}\times {\displaystyle \frac{3}{4}}\right]={\displaystyle \frac{1}{4}}$
Therefore, the probability of destroying the target in three shots,
$P\left(E_1\cup E_2\cup E_3\right)=P\left(E_1\right)+P\left(E_2\right)+P\left(E_3\right)$
Hence, substituting the known values in the above equation, we get
$\Rightarrow P\left(E_1\cup E_2\cup E_3\right)={\displaystyle \frac{1}{12}}+{\displaystyle \frac{7}{24}}+{\displaystyle \frac{1}{4}}$
Taking LCM and solving further we get,
$\Rightarrow P\left(E_1\cup E_2\cup E_3\right)={\displaystyle \frac{2+7+6}{24}}$
Adding the terms in the denominator, we get
$\Rightarrow P\left(E_1\cup E_2\cup E_3\right)={\displaystyle \frac{15}{24}}={\displaystyle \frac{5}{8}}$
Therefore, the probability of destroying the target in three shots is ${\displaystyle \frac{5}{8}}$
Hence, option C is the correct answer.

Note: We know that the probability of any event always lies between 0 and 1. If the probability of any event is 0 then, it is considered as an impossible event, as for example, if there are only blue balls in a bag, then, the probability of the ball taken out being red is 0 as this is impossible to happen. Similarly, if the probability of any event is 1, then, that particular event is a definite event. In this case only, the probability that the ball taken out is blue will always remain 1 as this is definite to happen.