Question

Two aeroplanes I and II bomb a target in succession. The probabilities of I and 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. $\frac{7}{22}$
• C. $0.2$
• D. $0.7$

Answer 1

Answer: (B)

$\frac{7}{22}$

Given: The probability of aeroplane I hitting correctly, $P\left(A\right)=0.3$, the probability of aeroplane II hitting correctly, $P\left(B\right)=0.2$

To find: the probability that the target is hit by the second plane

Sol: Probability of Aeroplane I hitting the target in first turn= $P\left(A\right)$

The probability of aeroplane II hitting the target in the first turn = $P\left(\overline{A}\right)\times P\left(B\right)$

Hence, the probability that the target is hit by the second plane =

$P\left(\overline{A}\right)\times P\left(B\right)+P\left(\overline{A}\right)\times P\left(\overline{B}\right)\times P\left(\overline{A}\right)\times P\left(B\right)+...⟹=0.7\times 0.2+0.7\times 0.8\times 0.7\times 0.2+...⟹=0.14[1+0.56+0.56\times 2+...]$

this is an geometric progression.

So the required probability becomes

$0.14\left[\frac{1}{1-0.56}\right]⟹=0.312=\frac{7}{22}$