Subject: Mathematics — Probability
Question: Let $|X|$ denote the number of elements in a set $X$.
Let $S = \{ 1,2,3,4,5,6\}$ be a sample space, where each element is
equally likely to occur. If $A$ and $B$ are independent events
associated with $S$, then the number of ordered pairs $(A,B)$ such that
$1 \leq |B| < |A|$, equals \_\_\_\_\_\_.
Answer: 422
Solution: Let
$$n(A) = a,\quad n(B) = b,\quad n(A \cap B) = c$$
$$1 \leq b < a$$
Also given that $A$ and $B$ are independent events,
$$P(A \cap B) = P(A)P(B)$$
$$\Rightarrow \frac{n(A \cap B)}{n(S)} = \frac{n(A)}{n(S)} \times \frac{n(B)}{n(S)}$$
$$\Rightarrow \frac{c}{6} = \frac{a}{6} \times \frac{b}{6} \Rightarrow ab = 6c$$
If $a = 6$, then $b = c = 5,4,3,2,1$ because $b < a$.
There is only one way to select all 6 elements of set $A$.
Number of ways of selecting 5, 4, 3, 2 or 1 elements in $B$ and
$A \cap B$ are
$$\binom{6}{5} + \binom{6}{4} + \binom{6}{3} + \binom{6}{2} + \binom{6}{1} = 2^{6} - 2 = 62$$
If $a = 5$, then
$$b = \frac{6c}{5},$$
which is not possible because if $c = 5$ then $b = 6$, while $b < a$.
If $a = 4$, then
$$b = \frac{6c}{4} = \frac{3c}{2},$$
which is possible because if $c = 2$ then $b = 3$.
2 elements in $A \cap B$ can be selected in $\binom{6}{2}$ ways. 2
additional elements in $A$ can be selected in $\binom{4}{2}$ ways. 1
additional element in $B$ can be selected in $\binom{2}{1}$ ways.
Therefore, number of ways for $a = 4$, $b = 3$, $c = 2$ are
$$\binom{6}{2} \times \binom{4}{2} \times \binom{2}{1} = 15 \times 6 \times 2 = 180$$
If $a = 3$, then
$$b = 2c \Rightarrow c = 1,\quad b = 2$$
which can be done in
$$\binom{6}{1} \times \binom{5}{1} \times \binom{4}{2} = 6 \times 5 \times 6 = 180\text{ ways.}$$
If $a = 2$, then $b = 3c$, which is not possible.
Therefore, total number of required ways
$$\text{Total} = 62 + 180 + 180 = 422.$$