Percentage of a failure given probability of failure Lets say I have a screen with $10^6$ pixels and the probability of a pixel failing is $p=10^{-6}$. What would be the percentage of unacceptable screens/failures if acceptable screens don't have more than 3 dead pixels?
Is this an example of a binomial distribution where the probability of an unacceptable screen is $1-P(1 \text{ failed pixel})-P(2 \text{ failed pixels})-P(3 \text{ failed pixels})$ ?
 A: Your count of defective pixels is $X \sim \mathsf{Binom}(n = 10^6, p =10^{-6}),$ which has $E(X) = np = 1.$ The exact probability of an unacceptable screen can be
computed using the binomial PDF formula or with software. From R statistical software the result is $P(X > 3) = 1 - [P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)] = 0.0190:$
1 - pbinom(3, 10^6, 10^-6)
##  0.0189881

Usually, $p$ is the binomial 'Success' probability, but you've defined the
events counted by $X$ as 'failed' pixels. Your method is OK, except you have
not accounted for $P(X = 0).$
Most practitioners would use an approximation, letting 
$X' \sim \mathsf{Pois}(\lambda = 1),$ where $E(X') = \lambda = 1.$
Then
$$P(X' > 3) = 1 - e^{-\lambda}(1 + \lambda/1! + \lambda^2/2! + \lambda^3/3!) = 0.9810.$$
 1 - ppois(3, 1)
 ##  0.01898816

 x = 0:3;  lam = 1;  pdf = exp(-lam)*lam^x/factorial(x);  1 - sum(pdf)
 ## 0.01898816

Because the 'expected' number of failures is 1, and the screen is 
acceptable if it has 0, 1, 2, or 3 bad pixels, it is not surprising that
the probability of an unacceptable screen is so low.
Below is a sketch of the PDF of $\mathsf{Pois}(1).$ The probability of
an unacceptable screen is represented by the sum of the heights of the
bars to the right of the vertical dotted red line. (At the scale of this
graph, the corresponding binomial probability bars would be indistinguishable
from the ones shown.)

