Question

. Break downs occur on a 20-years-old car with rate λ= 0.5 breakdowns/week. The owner of the car is planning to have a trip on his car for 2 weeks. What is the probability that there will be no breakdown on his car in the trip? [ The rate = ? per two weeks]

Answer 1

Answer: 0.3679

Explanation:

The formula for Poisson distribution :-

`P(x)=(e^(-\lambda)\lambda^(x))/(x!)`

Let x be the number of breakdowns.

Given : The rate of breakdown per week : 0.5

Then , for 2 weeks period the number of breakdowns =
`\lambda=0.5*2=1`

Then , the probability that there will be no breakdown on his car in the trip is given by :-

`P(x)=(e^(-1)1^(0))/(0!)=0.367879441171\approx0.3679`

Hence, the required probability : 0.3679