Answer to Question #196325 in Statistics and Probability for fahmida

Question

25. Suppose that in Bangladesh, three accidents occur in different factories per year on average. a. What is the probability of no accidents during a one-year period? 

b. What is the probability of at least two accidents during a one-year period? c. What is the probability of no accidents during the next six months?

Answer 1

Poisson Distribution formula

 P(x; μ) = (e^-μ) (μ^x) / x!

where

x=Poisson random variable (x)

μ=Average rate of success=3 (given in question)

 a. What is the probability of no accidents during a one-year period? 

sol.

Poisson random variable (x)=0

Average rate of success=3

 P(x; μ) = (e^-μ) (μ^x) / x!

 P(0; 3) = (e^-3) (3⁰) / 0!

= 0.04978706836

b. What is the probability of at least two accidents during a one-year period?

sol.

Poisson random variable (x)=2 at least

Average rate of success=3

in the case of at least variable , we find this as

1-[   P(0; 3) +    P(1; 3) ]

1-[  (e^-3) (3⁰) / 0! } +   (e^-3) (3¹) / 1! } ]

=0.80085

c. What is the probability of no accidents during the next six months?

sol.

Poisson random variable (x)=0

Average rate of success=3/2 (accidents in six months)

 P(x; μ) = (e^-μ) (μ^x) / x!

=0.22313