# Mixed Poisson Distribution Help

#### Asano

##### New Member
Question:
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!

#### Mean Joe

##### TS Contributor
You made this mistake: The distribution of Total Claimants is NOT Poisson with mean=28.

I'm not sure of a nice solution, but you can get T = Total Claimants <= 3 by:
a) 0 accidents
b) 1 accident
c) 2 accidents: 1 claimant-2 claimants, or 2 claimants-1 claimant
d) 3 accidents: 1 claimant-1claimaint-1 claimant

We then calculate the probability of the above scenarios:
a) P[0 accidents] = (12^0/0!) e^-12 = 6.1E-6
b) P[1 accident] = (12^1/1!) e^-12 = 7.4E-5
c) P[2 accidents] * P[1 claimant, and then 2 claimants] * 2 = (12^2/2!) e^-12 * (1/6) * (1/3) * 2 = 4.9E-5
d) P[3 accidents] * P[1 claimant, then 1 claimant, then 1 claimant] = (12^3/3!) e^-12 * (1/6)^3 = 8.2E-6

We then add up the above probabilities to finally determine P = 1.4E-4