$10$ indistinguishable balls are placed in $5$ distinguishable urns. Find the probability that exactly $2$ urns remain empty, given that not every box receives a ball.
So, this is a conditional probability that can be written as $\text{P(exactly $2$ urns remain empty/at least $1$ box remains empty)} = P(A|B) = \dfrac{P(A \cap B)}{P(B)}$
To choose the $2$ urns that remain empty we have $\displaystyle \binom{5}2$ ways and we have $\displaystyle \binom{7+3-1}{3-1}$ ways to put a ball in every other urn.
But I don't know how to calculate the above probability. Any help please? Thank you
Number of ways to distribute all balls in the urns is clearly given by $5^{10}$.
Now suppose if $a_1$ balls go in the first urn, $a_2$ balls go in the second urn and so on and $a_5$ balls go in the fifth urn , we have $$a_1+a_2+a_3+a_4+a_5=10$$
whose number of solutions by stars and bars is given by $\displaystyle \binom{10+5-1}{5-1}=\binom{14}{4}$.
Now suppose no urns remain empty, so we force the condition $a_i \geq 1 $ for all $i$ where $1\le i \le 5$.
So if we assume $a_1=b_1+1, a_2=b_2+1$ and so on so that we have $b_i \geq 0$ for all $i$.
So we have our new equation as $$b_1+b_2+b_3+b_4+b_5=5$$
whose number of solutions by stars and bars is given by $\displaystyle \binom{5+5-1}{5-1}=\binom{9}{4}$.
So this was the number of ways of distributing the balls so that no urn remains empty, so that means the number of ways in which atleast one urn remains empty is $\displaystyle \binom{14}{4}-\binom{9}{4}$.
Therefore the probability that atleast one urn remains empty is $$\displaystyle \dfrac{\binom{14}{4}-\binom{9}{4}}{5^{10}}$$
For computing the probability that exactly $2$ urns remain empty, let $a_4=a_5=0$ and force $a_1,a_2,a_3 \geq 1$, so by similar reasoning we have the equation $$b_1+b_2+b_3=7$$
whose solution by stars and bars is given by $\displaystyle \binom{7+3-1}{3-1}= \binom{9}{2}$. But we could let any two $a_i's$ to be $0$, so we multiply by $\displaystyle \binom{5}{2}=10$, so it's probability becomes $$\dfrac{\binom{5}{2}\binom{9}{2}}{5^{10}}$$
So by dividing the above two probabilities, we get out final answer as $\displaystyle \dfrac{\binom{5}{2}\binom{9}{2}}{\binom{14}{4}-\binom{9}{4}}=\dfrac{72}{175}$.