Question

The probability of a bomb hitting a bridge is $\frac{1}{2}$ and two direct hits are needed to destroy it. The least number of bombs required so that the probability of the bridge being destroyed is greater than $0.9$ is

• A. $8$
• B. $7$
• C. $6$
• D. $9$

Answer 1

The correct option is B

$7$

Explanation for the correct option:

Assume that $“n”$ be the least number of bombs required, and $X$ be the number of bombs required to hit the bridge.

General form of binomial distribution for any random variable $X$ is $\mathbf{P}\mathbf{(}\mathbf{x}\mathbf{:}\mathbf{n}\mathbf{,}\mathbf{p}\mathbf{)}\mathbf{=}\left(\frac{n}{x}\right){\mathbf{p}}^{\mathbf{x}}{\mathbf{q}}^{\mathbf{n}\mathbf{-}\mathbf{x}}\mathbf{=}\frac{\mathbf{n}\mathbf{!}}{\mathbf{(}\mathbf{n}\mathbf{-}\mathbf{x}\mathbf{)}\mathbf{!}\mathbf{x}\mathbf{!}}{\mathbf{p}}^{\mathbf{x}}{\mathbf{q}}^{\mathbf{n}\mathbf{-}\mathbf{x}}$

where $P=$binomial probability, $x=$ number of times for a specific outcome within n trials, $n=$number of trials, $p=$ probability of success on a single trial, $q=$probability of failure on a single trial and $\frac{n}{x}=$number of combinations.

Here $X$ follows the binomial distribution with $n=\frac{1}{2}$ and $p=\frac{1}{2}$

$P(X\ge 2)>0.9$

It means that,

$$\begin{gathered} 1-P(X<2)>0.9 \\ P(X=0)+P(X=1)<0.1 \\ {}^{n}C_0{\left(\frac{1}{2}\right)}^n+{}^{n}C_1\left(\frac{1}{2}\right){\left(\frac{1}{2}\right)}^{n-1}<0.1 \end{gathered}$$

$10(n+1)<2^n,$ which gives $n\ge 7.$

Therefore the least number of bombs required are $7.$

Hence the correct option is option(B)