Q. 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) 0.14
C) 0.32
D) 0.7
Explanation
(c)
Let A and B are the Ist and IInd aeroplane hit the target respectively and their corresponding probabilities are
$P(A) = 0.3\ and\ P(B) = 0.2$
$\Rightarrow \ P\left( \overline{A} \right) = 0.7\ and\ P\left( \overline{B} \right) = 0.8$
$\therefore$ Required probability
$= P\left( \overline{A} \right)P(B) + P\left( \overline{A} \right)P\left( \overline{B} \right)P\left( \overline{A} \right)P(B) + ...$
$= (0.7)(0.2) + (0.7)(0.8)(0.7)(0.2) +$
$(0.7)(0.8)(0.7)(0.8)(0.7)(0.2) + ...$
$= 0.14\left\lbrack 1 + (0.56) + (0.56)^{2} + ... \right\rbrack$
$= 0.14\left( \frac{1}{1 - 0.56} \right) = 0.32$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.