The number of accidents follows a Poisson distribution with mean 12. Each accident generates 1, 2 or 3 claimants with probabilities 1/6,1/3,1/2, respectively.

Calculate the probability that there are at most 3 claimants.

Attempt:

\(E(N) = 12[\frac{1}{6}+2*\frac{1}{3}+3*\frac{1}{2}] = 12*\frac{7}{3}=28\)

Step 2:

P(N≤3)= \(\sum_{x=0}^3 \frac{28^x e^{-28}}{x!} = 2.8*10^{-9} \)

That's the answer I got, but the number seems to be too small in value. Did I do anything wrong? Thank you!