Question

A home security system is designed to have a 99% reliability rate. Suppose that nine homes equipped with this system experience an attempted burglary. Find the probability that least one of the alarms is triggered.

Answer 1

Answer:

100% probability that least one of the alarms is triggered.

Step-by-step explanation:

For each alarm, there are only two possible outcomes. Either they are triggered, or they are not. The probability of an alarm being triggered is independent from other alarms. So 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.

A home security system is designed to have a 99% reliability rate.

This means that $p = 0.99$

Suppose that nine homes equipped with this system experience an attempted burglary.

This means that $n = 9$

Find the probability that least one of the alarms is triggered.

This is $P(X \geq 1)$

We know that either no alarms are triggered, or at least one is. The sum of the probabilities of these events is decimal 1. So

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

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

In which

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

$P(X = 0) = C_{9,0}.(0.99)^{0}.(0.01)^{9} \cong 0$

1 - 0 = 1

100% probability that least one of the alarms is triggered.