Variance of combination of Poisson and binomial distributions The physical situation I am trying to understand is the detection of photons by a digital camera sensor. During an exposure, photons will hit a particular pixel and the number hitting the pixel can be represented by a Poisson distribution where both the mean and variance are given by $\lambda$.
If every photon was detected by the sensor then that would be the end of the problem, but in fact only a fraction of the photons are detected - the probability of detecting each photon can be given by $p$, which is usually referred to as the quantum efficiency (QE) of the sensor.
It seems immediately obvious that the mean number of photons that will be detected in this situation is given by $\lambda p$, but I cannot see how to work out the variance of this number. 
If a fixed number, $n$, of photons hit the sensor then the variance would be given by the binomial distribution variance of $np(1-p)$, but how does the variance of this combine with the underlying variance due to the Poisson distribution of the number of photons hitting the sensor?
 A: For your problem, consider that you have two Poisson random variables.
Detected photons $D \sim \mathsf{Pois}(p\lambda)$ and Undetected photons
$U \sim \mathsf{Pois}((1-p)\lambda).$ Their sum is
$T = D + U \sim \mathsf{Pois}(\lambda).$
Assuming independence, the variances compute properly:
$Var(T) = \lambda$ and $Var(D) + Var(U) = p\lambda = (1-p)\lambda = \lambda.$

There is a more general formulation, sometimes called 'random sum of random
variables', which you might want to google. Suppose there are $N \sim \mathsf{Pois}(\lambda=20)$ customers
a day each independently spending $X_i \sim \mathsf{Norm}(\mu = 100, \sigma=15)$ dollars.
Then the total receipts in a day are $S = \sum_{i=1}^N X_i.$
In this situation $$E(S) = E(N)E(X) = \lambda\mu = 2000$$ and
$$V(S) = E(N)V(X) + V(N)[E(X)]^2 = \lambda\sigma^2 + \lambda\mu^2
= 20(225) + 20(10000) = 204500,$$ so $SD(S) = 452.2168.$

If you want to look at your photon problem according to this mechanism,
then the number of detected photons is $D = \sum_{i=1}^N X_i,$ where
$N \sim \mathsf{Pois}(\lambda)$ and the $X_i \sim \mathsf{Bernoulli}(p)
\equiv \mathsf{Binom}(1, p).$ So that $E(D) = E(N)E(X) = \lambda p$
and $V(D) = E(N)V(X) + V(N)[E(X)]^2 = \lambda p(1-p) + \lambda p^2
= \lambda p$ as previously. Note particularly that the $X_i$ are Bernoulli.

Here is a simulation in R statistical software for a million pixels
each with $\lambda = 10$ and $p = 0.7.$
m = 10^6;  d = numeric(m)
lam = 10;  p = .7
for (i in 1:m) {
   n = rpois(1, lam)
   d[i] = rbinom(1, n, p) }
mean(d); var(d)
## 7.000906   # aprx E(D) = 7
## 7.003016   # aprx V(D) = 7

In the figure below, histogram bars show the simulated values of $D$ and
the red circles show the exact Poisson probabilities.

