A large bank vault has several automatic burglar alarms. The probability is 0.55 that a single alarm will detect a burglar. (a) How many such alarms should be used to be 99% certain that a burglar trying to enter is detected by at least one alarm?

6 alarms to be 99% certain that a burglar trying to enter is detected by at least one alarm

Step-by-step explanation:

For each alarm, there are only two possible outcomes. Either a theft is detected, or is not. The probability of an alarm detecting a theft is independent of others 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.

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

And p is the probability of X happening.

0.55 that a single alarm will detect a burglar.

This means that

(a) How many such alarms should be used to be 99% certain that a burglar trying to enter is detected by at least one alarm?

The probability that no alarms detect a burglar is P(X = 0). Either no alarms detect, or at least one does. We need to have a 99% probability that at least one does. So we need P(X = 0) = 1 - 0.99 = 0.01.

We do by trial and error, find n until P(X = 0) <= 0.01.

n = 1

n = 2

n = 3

n = 4

n = 5

n = 6

So we need 6 alarms to be 99% certain that a burglar trying to enter is detected by at least one alarm

step-by-step explanation: