Question

An article in Fire Technology, 2014 (50.3) studied the effectiveness of sprinklers in fire control by the number of sprinklers that activate correctly. The researchers estimate the probability of a sprinkler to activate correctly to be 0.7. Suppose that you are an inspector hired to write a safety report for a large ballroom with 10 sprinklers. Assume the sprinklers activate correctly or not independently.
Required:
a. What is the probability that all of the sprinklers will operate correctly in a fire?
b. What is the probability that at least 7 of the sprinklers will operate correctly in a fire?

Answer 1

Answer:

a) 0.0282 = 2.82% probability that all of the sprinklers will operate correctly in a fire

b) 0.6496 = 64.96% probability that at least 7 of the sprinklers will operate correctly in a fire.

Step-by-step explanation:

For each sprinkler, there are only two possible outcomes. Either they will operate correctly, or they will not. The sprinklers activate correctly or not independently, 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.

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

And p is the probability of X happening.

The researchers estimate the probability of a sprinkler to activate correctly to be 0.7.

This means that $p = 0.7$

10 sprinklers.

This means that $n = 10$

a. What is the probability that all of the sprinklers will operate correctly in a fire?

This is $P(X = 10)$. So

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

$P(X = 10) = C_{10,10}.(0.7)^{10}.(0.3)^{0} = 0.0282$

0.0282 = 2.82% probability that all of the sprinklers will operate correctly in a fire.

b. What is the probability that at least 7 of the sprinklers will operate correctly in a fire?

This is:

$P(X \geq 7) = P(X = 7) + P(X = 8) + P(X = 9) + P(X = 10)$

So

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

$P(X = 7) = C_{10,7}.(0.7)^{7}.(0.3)^{3} = 0.2668$

$P(X = 8) = C_{10,8}.(0.7)^{8}.(0.3)^{2} = 0.2335$

$P(X = 9) = C_{10,9}.(0.7)^{9}.(0.3)^{1}= 0.1211$

$P(X = 10) = C_{10,10}.(0.7)^{10}.(0.3)^{0} = 0.0282$

Then

$P(X \geq 7) = P(X = 7) + P(X = 8) + P(X = 9) + P(X = 10) = 0.2668 + 0.2335 + 0.1211 + 0.0282 = 0.6496$

0.6496 = 64.96% probability that at least 7 of the sprinklers will operate correctly in a fire.