Question

A bomber wants to destroy a bridge. Two bombs are sufficient to destroy it. If four bombs are dropped, what is the probability that it is destroyed, if the chance of a bomb hitting the target is $0.4$,
If the answer is $\frac{a}{b}$ (HCF of $a$ and $b$ is $1$), then find $b-a$?

Answer 1

Given,Bomber can destroy bridge with two bombs.
Probability of bomb hitting, $P\left(B\right)=0.4$
Probability of bomb not hitting, $P\left(NB\right)=0.6$
$4$ bombs are dropped,
Given, distribution is in the form of binomial proability distribution,
$(P+q{)}^n=n{C}_0{P}^n+n{C}_0{P}^{n-1}{q}^1+......+n{C}_n{q}^n$
Probability of bridge destroyed $=$ (Probability only two bomb hits the bridge) $+$ (Probability three bomb hits the bridge)(Probability four bombs hits the bridge)
$=P(X=2)+P(X=3)+P(X=4)$
$=4C_2\times \left(P\right(B){)}^2\times (P\left(NB\right){)}^2)+4C_3\times (P\left(B\right){)}^3\times \left(P\right(NB\left)\right)+4C_4\times \left(P\right(B){)}^4$
$=6\times {\left(\frac{2}{5}\right)}^2\times {\left(\frac{3}{5}\right)}^2+4\times {\left(\frac{2}{5}\right)}^3\times \left(\frac{3}{5}\right)+{\left(\frac{2}{5}\right)}^4$
$=\frac{(216+96+16)}{625}=\frac{328}{625}$
Compare with given $\frac{a}{b}$, we have
$\Rightarrow a=328,b=625$
$\Rightarrow b-a=297$