Question

Two aeroplanes 1 and 2 bomb a target in succession.The probability of 1 and 2 scoring a hit correctly are 0.3 and 0.2 respectively. The second plane will bomb only when 1st one misses the target.The probability that the target is hit by 2nd plane is?

• A. 0.40
• B. 0.36
• C. 0.32
• D. 0.7

Answer 1

The correct option is C

0.32

For 2nd plane to hit the target following event will happen:

Let $P\left(A\right)$= Probability that 1st plane hits the target: 0.3

$P\left(B\right)$= Probability that 2nd plane hits the target: -.2

$P\left(E\right)=P\left(A{)}^{\prime }P\right(B)+P(A{)}^{\prime }P\left(B{)}^{\prime }P\right(A{)}^{\prime }P\left(B\right)+P\left(A{)}^{\prime }P\right(B{)}^{\prime }P\left(A{)}^{\prime }P\right(B{)}^{\prime }P\left(A{)}^{\prime }P\right(B)................\infty$

This is an infinite GP with common ratio ${}^{\prime }{r}^{\prime }=P\left(A{)}^{\prime }P\right(B{)}^{\prime }$

$Sum=\frac{a}{1-r}$

$a=P\left(A{)}^{\prime }P\right(B)=0.7\times 0.2=0.14$

$r=0.7\times 0.8=0.56$

$P\left(E\right)=\frac{0.14}{1-0.56}$

$P\left(E\right)=\frac{0.14}{0.44}$

$APPROXIMATELY=0.32$