For safety reasons, 3 different alarm systems were installed in the vault containing the safety deposit boxes at a Beverly Hills bank. Each of the 3 systems detects theft with a probability of 0.88 independently of the others. The bank, obviously, is interested in the probability that when a theft occurs,at least one of the 3 systems will detect it. What is the probability that when a theft occurs, at least oneof the 3 systems will detect it? 0.9959 Your answer should be rounded to 5 decimal places.

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Answer:

0.99827 = 99.827% probability that when a theft occurs, at least one of the 3 systems will detect it.

Step-by-step explanation:

For each system, there are only two possible outcomes. Either it detects the theft, or it does not. The probability of a system detecting a theft is independent of other systems. 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.

[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]

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

[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]

And p is the probability of X happening.

Each of the 3 systems detects theft with a probability of 0.88 independently of the others.

This means that [tex]n = 3, p = 0.88[/tex]

What is the probability that when a theft occurs, at least oneof the 3 systems will detect it?

[tex]P(X \geq 1) = P(X = 1) + P(X = 2) + P(X = 3)[/tex]

[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]

[tex]P(X = 1) = C_{3,1}.(0.88)^{1}.(0.12)^{2} = 0.03802[/tex]

[tex]P(X = 2) = C_{3,2}.(0.88)^{2}.(0.12)^{1} = 0.27878[/tex]

[tex]P(X = 3) = C_{3,3}.(0.88)^{3}.(0.12)^{0} = 0.68147[/tex]

[tex]P(X \geq 1) = P(X = 1) + P(X = 2) + P(X = 3) = 0.03802 + 0.27878 + 0.68147 = 0.99827[/tex]

0.99827 = 99.827% probability that when a theft occurs, at least one of the 3 systems will detect it.