Question

A hospital has three emergency generators for use in case of a power failure. Each generator functions independently, and the manufacturer claims that the probability that each generator will function properly during a power failure is 0.95.

1. List all of the possible outcomes in the sample space, S, and find the probability associated with each outcome.

2. Define the random variable X = the number of generators that fail. Calculate the value of X for each outcome in the sample space and give the probability mass function for X. Use correct notation.

3. Find the mean and variance of X.

4. Suppose that a power failure occurs and all three generators fail. Do you think that there would be reason to doubt the manufacturer‘s claim? Use probability to justify your answer.

Answer 1

(a)

Let F denote the case when a generator fails during a power failure, and P denote the case when the generator functions properly during the power failure.

There will be : 2^3 = 8 possible outcomes. The possible outcomes in the sample space along with their probabilities are listed below:

(1) P P P , probability = 0.95^3 = 0.857

(2) P P F , probability = (0.95^2)*0.05 = 0.045

(3) P F P , probability = (0.95^2)*0.05 = 0.045

(4) F P P , probability = (0.95^2)*0.05 = 0.045

(5) P F F , probability = (0.05^2)*0.95 = 0.00237

(6) F P F , probability = (0.05^2)*0.95 = 0.00237

(7) F F P , probability = (0.05^2)*0.95 = 0.00237

(8) F F F , probability = 0.05^3 = 0.000125

(b)

The pmf for this random variable X is as shown below:

P(X=0) = 0.857

P(X=1) = 3*0.045 = 0.135

P(X=2) = 3*0.00237 = 0.00711

P(X=3) = 0.000125

(c)

Mean of X = E[X] = 0*0.857 + 1*0.135 + 2*0.00711 + 3*0.000125 = 0.1496

Var(X) = E[X²] - (E[X])²

E[X²] = 0*0.857 + 1*0.135 + 4*0.00711 + 9*0.000125 = 0.1645

So,

Var(X) = 0.1645 - (0.1496^2) = 0.142

(d)

Probability that all the three generators fail is 0.000125, which is extremely small. So if this happens, the claim of the manufacturer cannot be doubted, because this would have occured by chance.