Question

Let $\left|X\right|$denote the number of elements in set $X$. Let $S=\left\{1,2,3,4,5,6\right\}$ 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 $1<\left|B\right|<\left|A\right|$, equals.

Answer 1

Finding the number of ordered pairs $\left(A,B\right)$

Given sample space $S=\left\{1,2,3,4,5,6\right\}$

$\left|X\right|$ denote the number of elements in the set $X$

Now $A$ and $B$ are independent events, hence,

$$\begin{gathered} P\left(A\cap B\right)=P\left(A\right).P\left(B\right) \\ \Rightarrow \frac{\left|A\cap B\right|}{\left|S\right|}=\frac{\left|A\right|}{\left|S\right|}.\frac{\left|B\right|}{\left|S\right|} \\ \Rightarrow \left|S\right|\left|A\cap B\right|=\left|A\right|.\left|B\right|\left(1\right) \end{gathered}$$

Now because $\left|A\right|>\left|B>1\right|$ so the possible value of $\left|A\right|$ are $2,3,4$ or $6$

$\left|A\right|\ne 5$ because it cannot satisfy the equation $\left(1\right)$

If $\left|A\right|=2$ then $\left|B\right|=1$. This also does not satisfy the equation$\left(1\right)$.

If $\left|A\right|=3$ then $\left|B\right|=2$ and $\left|A\cap B\right|=1$, number of ways is equal to

${}^{6}C_3\times {}^{3}C_1\times {}^{3}C_1=180$

If $\left|A\right|=4$ then $\left|B\right|=3$ and $\left|A\cap B\right|=2$, number of ways is equal to

${}^{6}C_4\times {}^{4}C_2\times {}^{2}C_1=180$

If $\left|A\right|=6$ then $\left|B\right|$ can be any number between $1$ to $5$.Therefore the number of ways is equal to

${}^{6}C_6\left[{}^{6}C_1+{}^{6}C_2+....+{}^{6}C_5\right]=62$

Therefore, the total number of ways or the total number of ordered pair equals $180+180+62=422$.