Question

Military radar and missile detection systems are designed to warn a coutry of an enemy attack. Assume that a particular detection system has a 0.80 probability of detecting a missile attack. If two military radars are insalled in two different areas and they operate independently, the probability that at least one of them detect an enemy attack is

Answer 1

Answer:

0.96 = 96% probability that at least one of them detect an enemy attack.

Step-by-step explanation:

For each radar, there are only two possible outcomes. Either it detects the attack, or it does not. The missiles are operated independently, which means that we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

Assume that a particular detection system has a 0.80 probability of detecting a missile attack.

This means that $p = 0.8$

If two military radars are installed in two different areas and they operate independently, the probability that at least one of them detect an enemy attack is

This is $P(X \geq 1)$ when $n = 2$. So

$P(X \geq 1) = 1 - P(X = 0)$

In which

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{2,0}.(0.8)^{0}.(0.2)^{2} = 0.04$

$P(X \geq 1) = 1 - P(X = 0) = 1 - 0.04 = 0.96$

0.96 = 96% probability that at least one of them detect an enemy attack.