What is the probability that a ball lands in a bin? Consider the classic problem of throwing balls to bins.
As usual, balls are thrown uniformly at random, and independently from one another, to the bins.
Let $N,B$ denote the number of balls and bins respectively.
Contrarily to the classic problem, however, balls can be discarded.
Each ball is discarded with probability $\alpha$.
Alike for the throws, balls are discarded independently from each other.
For each bin $b_{i}$, let $S_{i}$ denote a random variable corresponding to the number of balls landing in $b_{i}$.
What is the value of $P\left\{S_{i} = 0\right\}$?
(Note: if all balls were tossed, $P\left\{S_{i} = 0\right\} = \left(1 - \frac{1}{B}\right)^{N}$)
 A: Consider a simpler problem where N = 1 (i.e. there is only a single ball), and a fixed bin $b_{i}$.
The probability that the ball is discarded is $\alpha$.
If the ball is not discarded, the probability that it lands in $b_{i}$ is $\frac{1}{B}$. On the other hand, if the ball is tossed, the probability that it does not land in $b_{i}$ is $1 - \frac{1}{B}$.
Now, we can compute the probability that the ball does not land in $b_{i}$:


*

*The ball can either be discarded (with probability $\alpha$).

*Or it was tossed but simply did not hit the bin (with probability $\left(1 - \alpha\right) . \left(1 - \frac{1}{B}\right)$).


Summing up, the probability that the ball does not target $b_{i}$ is $\alpha + \left(1 - \alpha \right) . \left(1 - \frac{1}{B}\right) =  \left(1 - \frac{1 - \alpha}{B}\right)$.
Now, we only have to generalize this result to an arbitrary number of balls, denoted by $N$.
Since throws are independent from each other, the probability that none (out of $N$) hits $b_{i}$ is: $$P\left\{S_{i} = 0\right\} = \left( 1 - \frac{1 - \alpha}{B}\right)^{N}$$
