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