Question

A home security system is designed to have a 90% reliability rate. Suppose that 6 home equipped with this system experience an attempted burglary. Find the probability that at least two of the alarms are triggered.

Answer 1

0.999945 = 99.9945% probability that at least two of the alarms are triggered.

Explanation:

For each alarm, there are only two possible outcomes. Either it is triggered, or it is not. The probability of an alarm being triggered is independent of any other alarm, which means that the binomial probability distribution is used 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) = (n!)/(x!(n-x)!)`

And p is the probability of X happening.

A home security system is designed to have a 90% reliability rate.

This means that
`p = 0.9`

Suppose that 6 home equipped with this system experience an attempted burglary.

This means that
`n = 6`

Find the probability that at least two of the alarms are triggered.

This is:

`P(X \geq 2) = 1 - P(X < 2)`

In which

`P(X < 2) = P(X = 0) + P(X = 1)`

`P(X = x) = C_(n,x).p^(x).(1-p)^(n-x)`

`P(X = 0) = C_(6,0).(0.9)^(0).(0.1)^(6) = 0.000001`

`P(X = 1) = C_(6,1).(0.9)^(1).(0.1)^(5) = 0.000054`

`P(X < 2) = P(X = 0) + P(X = 1) = 0.000001 + 0.000054 = 0.000055`

Then

`P(X \geq 2) = 1 - P(X < 2) = 1 - 0.000055 = 0.999945`

0.999945 = 99.9945% probability that at least two of the alarms are triggered.