independent Poisson processes probability

Question

I have been given this question to solve

The numbers of claims to an insurance company from smokers and nonsmokers follow independent Poisson processes. On average 4 claims from nonsmokers and 6 claims from smokers arrive every day independently of each other.

Given that 8 claims arrived in a day, what is the probability that 5 of them were from smokers?

so my attempt is:

$X_{t} :$ number of claims from non-smokers (with intensity $\lambda = 4$)

$Y_{t} :$ number of claims from smokers (with intensity $\alpha = 6$)

So from this I have

$Z_{t}:$ number of claims from smokers & non-smokers (with intensity $\lambda+\alpha = 10$)

So I have set the probability to be

$$\frac{P(X_{1}=3)P(Y_{1}=5)}{P(Z_{1}= 8)}$$

I'm not sure if this is correct, I'm uncertain if I need to include $X_{t}$ in the equation.

Answer 1

$P(Y_1=5|Z_1=8)$

$ = \frac{P(Y_1=5,\; Z_1=8)}{P(Z_1=8)}$ (by Baye's rule)

$ = \frac{P(Y_1=5, \; X_1=3)}{P(Z_1=8)}$ (since $Z_1=X_1+Y_1$)

$ = \frac{P(Y_1=5)P(X_1=3)}{P(Z_1=8)}$ (since $X_1$ and $Y_1$ are independent)

Since $X_1,Y_1$, and $Z_1$ are all Poisson, you can easily plug in the standard formula in the above equation