Question

Bob owns a car rental company that has a fleet of 10 cars. Each car battery has an Exponentially distributed lifespan with a mean of 2 years. At the beginning of last year (Jan. 1, 2019), Bob replaced all the batteries. a) (5 points) What was the expected number of Batteries that Bob had to replace last year? b) (5 points) What was the probability that Bob had to replace 4 Batteries in 2019? c) (5 points) Bob replaced 3 batteries in June and 1 battery in October. It is now (Jan 1. 2020) and all the batteries are still working. What is the probability that he will have to replace 3 batteries in the coming year?

Answer 1

a) The probability that ny of the 10 batteries fail before an year is computed here as:

$P(T < 1) = \int_{0}^{1}(1/2)e^{-t/2}dt = 1 - e^{-1/2} = 0.3935$

The expected number of batteries that need to be replaced within 1 year is computed here as:
= 10*P(T < 1) = 10*0.3935 = 3.93

Therefore 3.93 is the expected number of batteries that Bob had to replace the given year.

b) The probability that 4 batteries had to be replaced in 2019 is computed here as:

$= \binom{10}{4} 0.3935^4(1 - 0.3935)^6 = 0.2506$

Therefore 0.2506 is the required probability here.

c) Note that exponential distribution follows memoryless property here, therefore it does not matter what happened in 2019, the probability that he will have to replace 3 batteries in 2020 is computed using Binomial probability function here as:

10" 10.3935(1-0.39357 = 0.2207 3

Therefore 0.2207 is the required probability here.