Answer to Question #326295 in Statistics and Probability for Devipriya

Question

The probability of hitting an aircraft is 0.001 for each shot. 

How many shots should be fired so that the probability of 

hitting with two or more shots is above 0.95?

Answer 1

$P(X \ge2)=1-P(X\le1)$

$1-P(X=1)-P(X=0) \ge 0.95$

$P(X=k)=C^n_k0.001^k0.999^k$

k=0,1,2...n

So P(X=0)=0.999ⁿ and

$\mathsf P(X=1)=n \cdot 0.001\cdot0.999^{n-1}$

Thus just find the smallest n such that

$1- 0.999^{n}-n\cdot0.001\cdot 0.999^{n-1}\geq 0.95$

$0.05 \ge 0.999^{n-1}(1+0.001n)$

n=5000

$0.05 \ge 0.0067x6=0.04$