Iām struggling to wrap my head around a basic balls and bins question:
So say I have $12$ indistinguishable balls and $6$ distinguishable bins. $3$ balls are already in $3$ different bins. The next $9$ balls are put into the $6$ bins randomly. What is the probability that each bin has at least one ball?
I approached this problem by dividing the number of ways to put $6$ balls in $6$ bins by the number of ways to put $9$ balls in $6$ bins. Thus:
$$\frac {\binom {11}{6}}{\binom {14}{9}} = \frac {462}{2002}$$
What if we start with $2$ balls in one bin and $1$ ball in another bin instead of $3$ balls in $3$ different bins?
Then I get $\frac {\binom {10}{5}}{\binom {14}{9}} = \frac {252}{2002}$.
Is this right? I assumed that the probability would be higher.
Best Answer
The question seeks the probability that each bin has at least one ball. You should not be using stars and bars method to calculate probability. Treat each ball as if they were distinguishable.
Now we are given that $3$ balls are already there in $3$ different bins. So we are left with placing $9$ balls in $6$ bins and each ball is placed in one of the bins randomly. So sample space is $6^9$.
Say for example, bins $1, 2, 3$ have a ball each to start with. So favorable outcomes are those where bins $4, 5, 6$ receive at least a ball each from $9$ balls. The best way to go about it is to apply Principle of Inclusion Exclusion.
So desired probability is,
$ \displaystyle \left [6^9 - {3 \choose 1} 5^9 + {3 \choose 2} 4^9 - {3 \choose 3} 3^9 \right] / 6^9 = \frac{830845}{1679616}$