Poisson distribution question: Company XYZ provides a warranty on a product that it produces... 
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 attempt:
Let $Y$ be the RV denoting the amount of annual insurance policy payments to Company XYZ and $X$ be the number of annual warranty claims. Then,
$$Y=5000(X-1) \;\; \text{if}\;\; X>1$$
and $0$ otherwise.
$$\begin{align}
E[Y] &= 5000 \cdot E[X-1\mid X>1] \\
&= 5000 \cdot \sum_{x=2}^\infty (x-1) \cdot P[X=x-1]\\
&=5000 \cdot \sum_{x=1}^\infty (x) \cdot P[X=x]\\
&=5000[E(X)-0\cdot P[X=0]]\\
&=5000[E(X)]
\end{align}$$
Note that $P(X=0) = 0.6 \implies c= -\ln 0.6$. So, $E[Y]=5000[-\ln 0.6].$
This is not the correct answer, however. Can someone please point out where I went wrong?
 A: Number of claims is $X \sim \mathsf{Pois}(c),$ so
$P(X=0) = e^{-c} = 0.6,$ which implies $c = -\log_e(.6) = 0.511.$ So $E(X) = 0.5108.$
c =-log(.6); c
[1] 0.5108256

Because of XYZ can't collect on the first claim,
you can't multiply $c$ by \$5000 to get the average
annual payout as $5000c.$
There may be some years with only
one claim so XYZ collects nothing, and there  may be years with several valid claims so thqt
XYZ collects from the insurance company.
Here is a strictly computational method.
Technically, Poisson probabilities extend from $0$ to $\infty.$ But terms decay rapidly to 0, Let's use the total number of claims for illustration.
You can come
extremely close to $E(X)$ by summing only the first 100 terms of the infinite series,
k = 1:99;  s = sum(k*dpois(k, -log(.6)))
s    
[1] 0.5108256  # exactly E(X) to many places

Now we find the expected number $Y$ of claims with payouts. We seek $E(Y) = 0.1108$ so that we can give the expected annual payout of $5000\,E(Y)\approx \$554$.
We let the 100-vector j represent the number
of paid claims and the 1000-vector i be the number of claims.
j = c(0,0,1:98);  i = 0:99
  5000*s
[1] 0.1108256  # E(Y) = avg nr. payable claims
[1] 554.1281   # 5000E(Y) = avg annual payout


Note: in case you're wondering whether it is sufficient to sum the first 100 terms of the
series, the first 50 would have been more than enough:
j = c(0,0.1:48); i = 0:48
sum(j*dpois(i, -log(.6)))
[1] 0.1508256

Almost all of the probability of the distribution $\mathsf{Pois}(-log(.6))$ is below 6.
qpois(.9999, -log(.6))
[1] 5

