If $7$ balls are distributed among three urns, what is the probability that one urn remains empty?

Question

Suppose $7$ balls are distributed among $3$ urns. What is the probability that one urn remains empty?

No. of ways to distribute $7$ balls to $3$ urns=$ 7^3$

No. of ways to pick an empty urn = $3\choose 1$

No. of ways to place $7$ balls to $2$ urns = $7\choose 2$

$\therefore P(\text{one urn remains empty})= \frac{{7\choose2}{3\choose1}}{7^3}$

I am sure the answer is wrong. I am a total novice in combinatorics. Please bear with me!!

Answer 1

Your denominator is flipped. There are $3$ slots to put the $7$ balls so it should be $3^7.$ After choosing the two slots, there are $2^7$ ways to place these so that they all get placed in those two urns, but this is over-counting.

You can use the inclusion exclusion principle to combat this. Let $X$ denote the number of distinct urns that the balls get placed in.

The probability that they all get placed in $1$ or $2$ urns is

$$P(X\in\{1,2\})={3 \choose 2}\cdot\left(\frac{2}{3}\right)^7$$

since we choose $2$ urns for all the balls to get placed in and the probability that all $7$ balls get placed in these urns is $\left(\frac{2}{3}\right)^7.$ However, this includes the cases where all $7$ balls get placed in one urn so we must subtract these off.

The probability that they are all in one urn is

$$P(X=1)={3 \choose 1}\cdot\left(\frac{1}{3}\right)^7$$

All together,

$$P(X\in\{1,2\})-P(X=1)\approx0.174$$

R Simulation:

urn = c(1,2,3) # a number for each urn
u = replicate(10^7, length(unique(sample(urn,7,repl=T))))
mean(u == 2);

[1] 0.1728419

which agrees with our analytical solution.

Answer 2

It depends on whether or not balls are distinguishable from other balls and likewise whether urns are distinguishable from other urns.

If both balls and urns are distinguishable, then the problem can be thought of in terms of working with ternary numbers. Each trit's position can be thought of as a ball and the urn its value. There are $3^7$ 7 trit ternary numbers. The situation in which one urn is empty is represented by numbers containing only two values of trit. There are ${3\choose2}*2^7-3$ of these (7 bit binaries with 3 symbol choices for minus the 3 with all bits the same). The probability is as stated by Remy.

If urns are distinguishable and balls aren't, then this problem is similar to rearranging the letters in the word ${\bf{bbbbbbbuu}}$ in a certain way. We know that there are $\bf\frac{9!}{7!2!}$ ways to rearrange the letters. This number is also the number of ways to place the seven indistinguishable balls into 3 urns. By imagining gluing letters together and using the same counting technique we find that the total number of ways that one urn ends up with no balls is $\bf21$, thereby making the probability of one empty urn $\bf\frac{7}{12}$. To check this I made a table with the ball count in each urn.

Now, if both balls and urns are indistinguishable, then the probability is $\bf\frac{1}{2}$. This answer is a bit hard to believe, but making a simple table will be enough to show it is true.