Question

Two aeroplanes $\text{I}$ and $\text{II}$ bomb a target in succession. The probabilities of $\text{I}$ and $\text{II}$ scoring a hit correctly are $0.3$ and $0.2$, respectively. The second plane will bomb only if the first misses the target. The probability that the target is hit by the second plane, is

• A. $0.06$
• B. $0.14$
• C. $0.32$
• D. $0.7$

Answer 1

The correct option is C

$0.32$

Find the required probability:

In the question, it is given that the probability of hitting a target by a bomb by two aeroplanes $\text{I}$ and $\text{II}$ is $0.3$ and $0.2$, respectively.

So, the probability of not hitting a target by a bomb by two aeroplanes $\text{I}$ and $\text{II}$ is $0.7$ and $0.8$, respectively.

So, The probability $P$ that the target is hit by the second plane is given by $P=(0.7)(0.2)+(0.7)(0.8)(0.7)(0.2)+(0.7)(0.8)(0.7)(0.8)(0.7)(0.2)+....$.

Take $(0.7)(0.2)$ common from the right-hand side.

$P=(0.7)(0.2)\left[1+(0.7)(0.8)+(0.7)(0.8)(0.7)(0.8)+....\right]$

Since, the terms in the square bracket are in G.P.

Therefore,

$$\begin{gathered} P=(0.7)(0.2)\left[1+(0.7)(0.8)+(0.7)(0.8)(0.7)(0.8)+....\right] \\ \Rightarrow P=(0.14)[1+0.56+{(0.56)}^2] \\ \Rightarrow P=(0.14)\left[\frac{1}{(1-0.56)}\right] \\ \Rightarrow P=0.32 \end{gathered}$$

Therefore, the probability that the target is hit by the second plane, is $0.32$.

Hence, option (C) is the correct answer.