Problema Solution
At a cafeteria the customers arrive at an average of 0.3 per minute. The probability that a)exactly 2 customers arrive in a 10 minute span
b)2 or more customers arrive in a 10 minute span
c) one customer arrives in a 5 minute span and one customer arrives in the next minute span is
Answer provided by our tutors
We use a Poisson process with a parameter l*t, l*t = 0.3 × 10 = 3 in items a) and b)
a) P(2 customers in 10 min span) = e^(-0.3*10)(0.3*10)^2/2! = 0.224
b) P(2 or more in 10 min span) = 1 - P(0) - P(1) = 1 - e^(-0.3*10)(0.3*10)^0/0! - e^(-0.3*10)(0.3*10)^1/1! = 0.8
c) P(one customer in 5 min and one customer in the next min) = (e^(-0.3*5)(0.3*5)^1/1!) * (e^(-0.3*1)(0.3*1)^1/1!) = 0.074
e = 2.71828 is Euler's number.