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Two aeroplanes $I$ and $II$ bomb a target in succession. The probabilities of $l$ and $II$ scoring a hit correctlyare $0.3$ 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
$0.2$
$0.7$
$0.14$
$0.32$
Solution
Let the events, $A=$ Ist aeroplane hit the target
$\mathrm{B}=$ IInd aeroplane hit the target
And their corresponding probabilities are
$P(A)=0.3 \text { and } P(B)=0.2$
$\Rightarrow \quad P(\bar{A})=0.7 \quad$ and $\quad P(\bar{B})=0.8$
$\therefore$ Required probability $=P(\bar{A}) P(B)+P(\bar{A}) P(\bar{B}) P(\bar{A}) P(B)+$
$=(0.7)(0.2)+(0.7)(0.8)(0.7)(0.2)$$+(0.7)(0.8)(0.7)(0.8)(0.7)(02)+. . .$
$=0.14\left[1+(0.56)+(0.56)^{2}+\ldots .\right]$
$=0.14\left(\frac{1}{1-0.56}\right)=\frac{0.14}{0.44}=\frac{7}{22}=0.32$
