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. 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]

Respuesta :

JeanaShupp

Answer: 0.3679

Step-by-step explanation:

The formula for Poisson distribution  :-

[tex]P(x)=\dfrac{e^{-\lambda}\lambda^{x}}{x!}[/tex]

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 = [tex]\lambda=0.5\times2=1[/tex]

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

[tex]P(x)=\dfrac{e^{-1}1^{0}}{0!}=0.367879441171\approx0.3679[/tex]

Hence, the required probability : 0.3679

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