# Quod Erat Demonstrandum

## 2008/03/03

### Simple probability question, be careful in the way of counting

Filed under: Additional / Applied Mathematics,HKALE — johnmayhk @ 5:19 下午

Two black balls and thirteen white balls are randomly put into five identical boxes, so that each box contains three balls. Find the probability that the two balls are put in different boxes. (3 marks)

The above is a simple question from applied mathematics (II) past paper, the fastest way may be considering the following.

After placing a black ball, there are 14 possible places for placing the other, but, not letting black balls in the same box, the 2 positions in the same box as the first black balls should not be occupied; hence, there are 14 – 2 = 12 remaining positions can be used for placing the second black ball; thus, the required probability is $\frac{12}{14} = \frac{6}{7}$.

Alternatively, we may use $C_r^n$, we may think that there are 15 positions to place the two black balls, hence there are $C_2^{15}$ combinations for placing two black balls. For not letting two black balls being in the same box, we can, first, choose 2 boxes out of 5 boxes, with the number of ways $C_2^5$. Now, there are two boxes selected, time to place the black balls. The first black ball will be placed in the first box, don’t forget that there are 3 positions in each box, hence there are $C_1^3$ way to place. After placing the first black ball, similarly, there are 3 positions for the second ball to be placed in the second box with $C_1^3$ ways, hence the total number of favourable outcome is $C_2^5C_1^3C_1^3$; therefore the required probability is $\frac{C_2^5C_1^3C_1^3}{C_2^{15}} = \frac{6}{7}$.

Afterwards, a student, Wong, suggested the following.

The required probability = 1 – P(two black balls in the same box)

In calculating P(two black balls in the same box), he gave the number of favourable outcomes as

taking 1 box out of 5 boxes = = => $C_1^5$ ways;
placing the first black in the first position of the box = = => $C_1^3$ ways;
placing the second black in the second position of the box = = => $C_1^2$ ways;

hence total number of favoriable outcomes = $C_1^5C_1^3C_1^2$

and the answer obtained from the so-called solution is $1 - \frac{C_1^5C_1^3C_1^2}{C_2^{15}} = \frac{5}{7} \ne \frac{6}{7}$, why?

The problem is, two balls are indistinguishable. Are there $C_1^3C_1^2$ ways of placing two indistinguishable balls in the same box? It seems to be too many! Actually, there are only $\frac{C_1^3C_1^2}{2!} = 3$ ways instead. Or simply write $C_2^3$ to represent getting 2 out of 3 positions to place the two black balls.

It is always a problem in teaching the way of counting, may be it’s better to give “what’s wrong" examples to let students learn abstract concepts in a complete way.