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

Step-by-step explanation:

The formula for Poisson distribution  :-

$P(x)=\dfrac{e^{-\lambda}\lambda^{x}}{x!}$

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 = $\lambda=0.5\times2=1$

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

$P(x)=\dfrac{e^{-1}1^{0}}{0!}=0.367879441171\approx0.3679$

Hence, the required probability : 0.3679