**1. The problem statement, all variables and given/known data**

An actuary has discovered that policyholders are three times as likely to file two claims as to file four claims. If the number of claims filed has a Poisson distribution, what is the variance of the number of claims filed?

**2. Relevant equations[/]**

P(x)=e^(-lamda)*(lamda^x)/x! Var(x)=lamda

I know that P(2)=3P(4) and that P(2)=e^(-lamda)*(lamda^2)/2 and that P(4)=e^(-lamda)*(lamda^4)/4! So I set these equal to find lamda and that is the variance. The solution given to me is P(2)=e^(lamda)*(lamda^2)/2=3*e^(-lamda)*(lamda^4)/4!=3*P(4)*24*lamda^2=6*lamda^4 I don't understand the last two equalities. Any help? where do the 24*lamda^2 comefrom and how do they get 6*lamda^4?

Thanks

P(x)=e^(-lamda)*(lamda^x)/x! Var(x)=lamda

**3. The attempt at a solution**I know that P(2)=3P(4) and that P(2)=e^(-lamda)*(lamda^2)/2 and that P(4)=e^(-lamda)*(lamda^4)/4! So I set these equal to find lamda and that is the variance. The solution given to me is P(2)=e^(lamda)*(lamda^2)/2=3*e^(-lamda)*(lamda^4)/4!=3*P(4)*24*lamda^2=6*lamda^4 I don't understand the last two equalities. Any help? where do the 24*lamda^2 comefrom and how do they get 6*lamda^4?

Thanks