probability expected payment Company XYZ provides a warranty on a product that it produces. Each year, the number
of warranty claims follows a Poisson distribution with mean c. The probability that no
warranty claims are received in any given year is 0.60.
Company XYZ purchases an insurance policy that will reduce its overall warranty claim
payment costs. The insurance policy will pay nothing for the first warranty claim
received and 5000 for each claim thereafter until the end of the year.
Calculate the expected amount of annual insurance policy payments to Company XYZ.
My solution
y= payment by insurance company
$$y=\begin{cases}
{0} & \text{if } x=0\\
5000(x-1) & \text{if }  x\geq1,\\ \end{cases}$$
c=0.5018
$p(x=0)=\frac{(e^-c)*(c^0)}{0!}=0.60$ so $c=0.5108$ so the mean of x is 0.5108 so $E[x]=0.5108$
$E[y]=\sum 0*p(x=0)+5000(x-1)*p(x\geq1)=0*06+5000(x-1)*0.4=2000E[x]-2000=-976$
Correct answer is 554,I would like to know if my function for payment is correct and what would be my next step.

We know: $X\sim\mathcal {Pois}(c)$ and $\mathsf P(X=0)=0.60$.  
Since the first fact means that $\mathsf P(X=x)~=~\dfrac{c^x\mathsf e^{-cx}}{x!}\mathbf 1_{x\in\Bbb N}$ , we can easily calculate $c$ knowing the second fact.  

We know $Y := 5000(X-1)^+~$  which is $~Y=5000\max(X-1,0)$
Then $\mathsf E(Y) = 5000~\mathsf E(X-1\mid X\geq 1)~\mathsf P(X\geq 1) \color{silver}{+ \require{cancel}\cancel{0~\mathsf P(X=0)}}$
If only the Poisson distribution had some convenient property that allowed us to easily find this conditional expectation without messy summation.   Hmm...
