A parking lot has two entrances Car arrives at entrance I ac

A parking lot has two entrances. Car arrives at entrance I according to a Poisson distribution at an avera of three per hour and at entrance II according to a Poisson distribution at an average of four per hour. What is the probability that a total of three cars will arrive at the parking lot in a given hour? Assuming that the numbers of cars arriving at the two entrances are independent)/

Solution

The Poisson formula is P(x=k)= e(-u)u^(k)/k! where u is the mean of the distribution.
So the probabilities for entrance 1 are P(x=0) = 0.0498, P(x=1)= 0.1494, P(x=2)= 0.2240, P(x=3)= 0.2240
For entrance 2 they are, P(x=0) =0.0183, P(x=1)= 0.0733, P(x=2)= 0.1465,P(x=3)= 0.1954
So the possible combinations are (0,3), (1,2),(2,2), and (3.0).
(0,3)= 0.0498*0.1954= =0.0097
(1,2)= 0.1494*0.1465= 0.0219
(2,1)= 0.2240* 0.0733= 0.0164
(3,0)= 0.2240*0.0183= .0041
Total probability is 0.0521