Saw this discussion and I can't understand the answer.
Let's say there are 90 unique toys, and a child has a specific one that he wants to get. He opens 225 boxes of cereal to receive this specific toy. After he has opened them all he received all toys except the one toy that he was interested in. What is the probability of this?
I saw the formula
$\sum_{k=0}^{89} (-1)^k Binomial[89,k] (1- \frac{k+1}{90})^{225} \approx 0.0000202174$
but I don't understand how we got this formula to calculate the probability of this. First of all, is Binomial(n,p) here the binomial coefficients or is it the binomial probability mass function, or something else? Can someone help me understand how we got this formula?
I saw the inclusion-exclusion principle get mentioned but don't really understand how they relate.
Binomial (89,k) does indeed correspond to $\binom {89} k.$
We start with an easier problem: what is the probability of not getting $k$ specific toys in $225$ boxes? Since there are $90$ possible toys and each box is independent, this is just $(1-\frac{k}{90})^{225},$ the probability of not getting one of our specific k toys $225$ times in a row.
Now, we don't want just the probability of not getting one specific toy, we also want the probability of getting every other toy too. This is where inclusion exclusion comes in.
The first term in the sum is: $$ (1-\frac 1 {90})^{225}$$ Which is exactly the $k=1$ case above, ie the probability we don't get the toy we want. But this over counts! It includes events like us getting only $88$ or $5$ of the other toys instead of all $89$.
So we correct with the next term:
$$- 89 (1-\frac 2 {90})^{225}$$
Which is exactly removing all events where we don't get $2$ particular toys, one of which is the one we wanted. We multiply by $89,$ the number of second toys we might also not get. But this time we've over corrected: we have double counted any event where we didn't get $3$ of the toys. So we must add these back.
This repeats for some time. The $k$th term is the probability of not getting $k+1$ toys, with one of those being the toy we wanted.