The question in my assignment is,
Suppose that N, the number of cars abandoned weekly on a certain highway, is a Poisson random variable with parameter 2.2, i.e.
P(N = n) = (n = 0, 1, 2, . . .) $\frac{e^-2.2((2.2)^n)}{n!}$ (So poisson distribution!) exp
Find the Probability that there is:
(a) no abandoned cars in the next week [5] (b) Find the probability that at least 3 abandoned cars in the next week. given that 1 abandoned cars in the next week.
I feel like (b) is ambiguous but my solution is,
P(N>=3|N>=1)= P(N>=3)/P(N>=1)
= (1-P(N<3))/(1-P(N>=1))
= 0.4243
If anyone has any thoughts if I went wrong it's be much appreciated! Thank you!