A garage has 2 vans and 3 cars, which can be hired out for a day at a time. Requests for the hire of a van follow a Poisson distribution with a mean of 1.5 requests per day and requests for the hire of a car follow an independent Poisson distribution with a mean of 4 requests per day.


Find the probability that, on any particular day, there is at least one request for a van and at least two requests for a car, given that there are a total of 4 requests on that day. (The answer is 0.286, apparently.)


My method:


Let V be number of requests for a van per day. Then V ~ Po (1.5).
Let C be number of requests for a car per day. Then C ~ Po (4).
Let T be total number of requests per day. Then T ~ Po (5.5).


Required probability = \frac {P(V=2)P(C=2) + P(V=1)P(C=3)}{P(T=4)}
= 0.656.


Can someone explain where I went wrong? Thanks!