Question

A missile guidance system has 5 fail-safe components. The probability of each failing is 0.05. Find the probability that more than 2 will fail.

Answer 1

We will use binomial probability for this.

Since we have 5 fail-safe components, the has 6 terms, 0 to 5.

We want only the term for 3, 4 and 5 fails.

The term for 3 fails and 2 successes is:

$P(3)=\frac{5!}{3!(5-3)!}0.95^20.05^3=10\cdot0.9025\cdot0.000125=0.001128125$

The term for 4 fails and 1 success is:

$P(4)=\frac{5!}{4!(5-4)!}0.95^10.05^4=5\cdot0.95\cdot0.00000625=0.0000296875$

The term for 5 fails and no success is:

$P(5)=\frac{5!}{5!(5-5)!}0.95^00.05^5=1\cdot1\cdot0.0000003125=0.0000003125$

The sum is:

$P(3)+P(4)+P(5)=0.001158125=0.1158125\%$