Question

An engine manufacturer has observed failures with their product. In the last 6 years, they have experienced 2 engine failures one year, 3 engine failures in two consecutive years, 4 engine failures one year, and 5 engine failures in each of the preceding two years (two in the last month, alone). The CEO of the engine manufacturer wants to reassure customers. He promises that in the next month, there will not be multiple engine failures (no more than 1 failure). Assume the probability of failure is exponential.

Question: The CEO asks you to calculate the probability that they are correct: What is the probability there will be no more than a single engine failure in the next month?

Note: Making promises to reassure customers and then asking to hear the numbers. This would never happen in the real-world, by the way.

Answer 1

Solution:

Let us tabulate the given data from the question to calculate the mean failure since we know the probability of failure is exponential.

Year	Number of engine failures
1	2
2	3
3	3
4	4
5	5
6	5
Mean failure	3.667

We need to find the probability that only 1 or 0 engine fails in the next month.

Probability of exponential distribution , P(x)= * e^(-.x) where x is number of failure.

We know mean=3.667=1/( from the above table)

So =1/3.667=0.2727

Let us substitute the value of in the below equation.

P(x)= * e^(-.x)=0.2727* e^(-0.2727x)

Probability that '0' engine fails P(x=0)=0.2727* e^(-0.2727*0)=0.2727

Probability that '1' engine fails P(x=1)=0.2727* e^(-0.2727*1)=0.2076

So,P(x=0)+P(x=1)=0.2727+0.2076=0.4803