Probability of finding 1 ball in N boxes choosing randomly n boxes

Question

Let’s say there’s 1 ball hidden in N boxes. You can open $n<N$ boxes. What’s the probability of finding the ball?

My approach: the way of choosing $n$ from $N$ boxes $$ \binom{N}{n}$$ Choosing the one that has the ball there are $N-1$ boxes left and $n-1$ to pick so $$\binom{N-1}{n-1}$$ ways to do it. Then the probability is $$\frac{\binom{N-1}{n-1}}{\binom{N}{n}}=\frac{n}{N}$$ Probability is not my forte. I think I’m wrong… am I?

Edit: I wasn't wrong thanks to all, but the logic would be the same if there were $k<n$ balls hidden? (one ball per box only)

Answer 1

Your answer is right. There's a couple ways of thinking about it; the comments on your post talk about a simple equivalent problem (think about why there's a bijection between the 2 ways of counting!).

You could even just brute force it. Either you find it on the first try, the second try, ..., the nth try. This probability comes to

$\frac{1}{N} + \frac{N-1}{N} \cdot \frac{1}{N-1} + \frac{N-1}{N} \cdot \frac{N-2}{N-1} \cdot \frac{1}{N-2} + ... + \frac{N-1}{N} \cdot \frac{N-2}{N-1} \cdot ... \cdot \frac{N-(n-1)}{N-(n-2)} \cdot \frac{1}{N-(n-1)}$

and that's equal to $\frac{1}{N} + \frac{1}{N} + ... + \frac{1}{N} = \frac{n}{N}.$

Your original answer also works well and is quite elegant :)

Answer 2

Since questions were raised over the comment I made, and you also tried to gauge my intention of how to directly arrive at an answer, I feel I should clarify.

Your answer is fine, and the method you have used is very useful, specially for more difficult questions.

The model I would have used here is that I have $n$ tickets in a lottery with $N$ total tickets, and only $1$ prize.

Then, without further thought, I could write $Pr = \large\frac{n}{N}$