We have $n$ boxes (numbered from $1$ to $n$) and $n$ balls (also numbered from $1$ to $n$). We put the balls in the boxes randomly, so that each $n^n$ outcome is equally likely.
$(a)$ What is the probability that the first two boxes will be empty?
$(b)$ What is the probability that there will be an empty box?
$(c)$ What is the probability that the first two balls end up in the same box?
For part $(a),$ I computed that the probability is equal to $\large \left( \frac{n-2}{n} \right)^n$, since for the first ball we have a probability of the first two boxes ending up empty being equal to $\large \frac{n-2}{n},$ and the same for the second ball, and so on. So that to find the probability of the first two boxes ending up empty we have to multiply the above.
For part $(b)$, I have computed that the probability of there being an empty box is equal to $\large \left( \frac{n-1}{n} \right)^n,$ since the probability of a ball landing in any but one specific box is equal to $\large\frac{n-1}{n}.$ And we multiply the above in order to find the desired probability.
For part $(c)$, I have no idea how to proceed. Could you provide a hint, and a hint only?
Is the above reasoning correct? If not, could you point me in the right direction? Thank you in advance.
Best Answer
Part (a) looks good.
But (b) does not. Look at the complementary event, that there are no empty boxes. How many ways can that happen?