An article in Fire Technology, 2014 (50.3) studied the effectiveness of sprinklers in firecontrol 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?
c. What is the minimum number of sprinklers needed so that the probability that at least one operates correctly is at least 0.98?

Answer:

0.028 ; 0.649 ; 4

Step-by-step explanation:

P(correctly activation), p = 0.7

Number of sprinklers, n = 10

q = 1 - p = 1 - 0.7 = 0.3

Using the binomial probability relation :

P(x =x) = nCx * p^x * (1 - p)^(n - x)

Probability that all sprinklers activate correctly;

P(x = 10) = 10C10 * 0.7^10 * 0.3^0

P(x = 10) = 1 * 0.0282475249 * 1

P(x = 10) = 0.028

Probability that atleast 7 will operate correctly :

P(x ≥ 7) = p(x = 7) + p(x = 8) + p(x = 9) + p(x = 10)

P(x ≥ 7) = 0.267 + 0.233 + 0.121 + 0.028

P(x ≥ 7) = 0.649

3.)

Probability of atleast 1 operates :

P(x ≥ 1) = 0.98

1 - probability of 0 operates

1 - p(x =0)

P(x = 0) = nC0 * 0.7^0 * 0.3^n = 0.02

Recall :

nC0 = 1 ;

1 - p(x = 0)

P(x = 0) = 1 * 1 * 0.3^n = 0.02

0.3^n = 0.02 - - - (1)

0.3^3.24 = 0.02 - - - (2)

Comparing (1) and (2)

n = 3.24

Since, n cannot be a fraction ;

Then n is rounded to the next whole number = 4

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