A machine wraps and twists the ends of a wrapper for pepperm

A machine wraps and twists the ends of a wrapper for peppermint candy at a rate of 100 per minute. However, the machine malfunctions on the average of 1.2 per minute. Assume that the malfunctions are a Poisson distribution and let x denote the number of candies which are not wrapped due to the malfunction. What is the probability that between zero and four candies will be unwrapped in a given minute? (Round your answer(s) to 3 decimal places.)

Solution

Given X follows Poisson distribution with mean=1.2 per minute

P(X=x)=(1.2^x)*exp(-1.2)/x! for x=0,1,2,...

So the probability is

P(0<=X<=4) = P(X=0)+P(X=1)+P(X=2)+P(X=3)+P(X=4)

=(1.2^0)*exp(-1.2)/1+....+(1.2^4)*exp(-1.2)/4!

=0.992