Question

If electricity power failures occur according to a Poisson distribution with an average of 3 failures every twenty weeks, calculate the probability that there will not be more than one failure during a particular week.

Answer 1

Let X be the Poisson random variable. It is given that mean λ = 3/20 = 0.15 

The Poisson probability law, giving x failures per week is given by,

P(X = x) = (e^-λλ^x)/x! 

= (e^-0.15(0.15)x)/x!, 

x = 0,1,2,3,..........

Hence probability that there will not be more than one failure is given by P (X ≤ 1) 

= P(X = 0) + P(X = 1) 

= e^-0.15 [1 + 0.15] 

= e^-0.15 (1.15) 

= (0.8607) (1.15) 

= 0.98981