The Problem : If $n$ identical balls are placed at random into $n$
cells, find the probability that exactly one cell remains empty.
A similar question was asked on this website : Probability of n balls in n cells, one remaining empty
So here's what I have done :
Let $c_i$ denote the number of balls in $i$th cell
Then
$c_1+c_2+\cdots+c_n=n \tag 1 $
So
Total number of outcomes $=$ Number of Solutions of equation $(1)$ [where $c_i =0$ is possible] $\displaystyle = {2n-1\choose n-1}$
Total number of favourable outcomes $=$ Number of Solutions of equation $(1)$ [ where one of $c_i=0$ and the rest are non-zero]$\displaystyle = \overbrace{n\choose 1}^{\text{Choosing } c_i =0}\underbrace{n-1\choose n-2}_{\text{Solutions to remaining}}$
Using this we get
$\displaystyle P(A)=\frac{n(n-1)}{2n-1\choose n-1}$
But some say this is wrong as all the solutions(which are the outcomes) aren't equally likely
So my question is why aren't they equally likely?
Best Answer
Consider $n=2$. The probability that both balls go in the left bin is $1/4$, because the only way it can happen is (left, left). But the probability that there is one ball in each bin is $2/4$, because there are two sequences which produce this outcome, (left, right) and (right, left). For the one-ine-each-bin outcome, no matter where you throw the first ball, the second ball always has a chance to land in a different bin; this additional freedom increases the probability of success.
In general, whenever you have several independent random variables, like die rolls or ball tosses, then you need to treat each item as distinct for counting purposes. Thing of this as a consequence of the definition of independence.