# If six different balls are placed in three different boxes, what is the probability that exactly two balls are placed in the first box?

If six different balls are placed in three different boxes, what is the probability that exactly two balls are placed in the first box?

My attempt: $k_1+k_2+k_3=6$, $k_i$ - quantity of balls in $i$ box. $k_1=2$ hence there are 4 sets for $k_2, k_3:$ $\{0,4\},\{1,3\},\{3,1\},\{4,0\}$.
Adding set $\{2,2\}$ we get all combinations when there are 2 balls in the first box $\Rightarrow$ possibility $= \frac{4}{5}$?

• Are you asking what the probability is that two balls are placed in the first box if six balls are distributed to three boxes? – N. F. Taussig Sep 30 '17 at 0:37
• yes, I do. English isn't my native language, hence sometime my states are bad to uderstand – ioleg19029700 Sep 30 '17 at 0:39
• yes. Let's say they have numbers from 1 to 6 – ioleg19029700 Sep 30 '17 at 0:41

There are $3^6$ ways to distribute six balls to three boxes since there are three choices for each of the six balls.
There are $\binom{6}{2}$ ways to select exactly two of the six balls to be placed in the first box and $2$ ways to select the box in which each of the four remaining balls is placed. Hence, there are $$\binom{6}{2}2^4$$ favorable cases.
Therefore, the probability that exactly two balls are placed in the first box when six balls are distributed to three different boxes is $$\frac{\binom{6}{2}2^4}{3^6}$$
Note: You attempted to count cases by considering how many balls are placed in each box. However, the $$\binom{6 + 3 - 1}{3 - 1} = \binom{8}{2}$$ solutions to the equation $$k_1 + k_2 + k_3 = 6$$ in the nonnegative integers are not equally likely to occur. There is only one way to place all six balls in the first box. However, there are $$\binom{6}{2}\binom{4}{2}\binom{2}{2} = 90$$ ways to place two balls in each box.
• +1 nice answerTaussig as usual...I did this at the numerator $\binom{6}{2}(\binom{4}{0}+\binom{4}{1}+\binom{4}{2}+\binom{4}{3}+\binom{4}{4})=2^4\binom{6}{2}$ – Isham Sep 30 '17 at 1:33
• @Isham That works. The number of subsets of a set with $n$ elements is $2^n$ since each element is either included or not included in a subset. Hence, a set with four elements has $2^4$ subsets. The number of subsets of size $k$ of a set with $n$ elements is $\binom{n}{k}$. Since the number of elements in a subset of a four-element set ranges from $0$ to $4$, $$\binom{4}{0} + \binom{4}{1} + \binom{4}{2} + \binom{4}{3} + \binom{4}{4} = 2^4$$ – N. F. Taussig Sep 30 '17 at 1:39