Number of Ordered Pairs of Independent Events in Finite Sample Space

To find the number of ordered pairs of independent events $(A, B)$ associated with the sample space $S = \{1, 2, 3, 4, 5, 6\}$, let $|A| = a$ and $|B| = b$.

Since all elementary outcomes in $S$ are equally likely: $P(A) = \frac{a}{6}, \quad P(B) = \frac{b}{6}, \quad \text{and} \quad P(A \cap B) = \frac{|A \cap B|}{6}$

For events $A$ and $B$ to be independent, we have: $P(A \cap B) = P(A) P(B) \implies \frac{|A \cap B|}{6} = \frac{a}{6} \cdot \frac{b}{6} \implies |A \cap B| = \frac{ab}{6}$

Since $|A \cap B|$ must be an integer, $ab$ must be a multiple of $6$. We are given the condition: $1 \le b < a \le 6$

We analyze all possible pairs $(a, b)$ satisfying these conditions:

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Case 1: $a = 6$

Here, $A = S$. For any $b \in \{1, 2, 3, 4, 5\}$: $|A \cap B| = |S \cap B| = |B| = b = \frac{6b}{6}$ Thus, every subset $B$ of size $b$ is independent with $A = S$.

• The number of ways to choose $A$ is $\binom{6}{6} = 1$.
• The number of ways to choose $B$ is: $\binom{6}{1} + \binom{6}{2} + \binom{6}{3} + \binom{6}{4} + \binom{6}{5} = 6 + 15 + 20 + 15 + 6 = 62$
• Total pairs for this case: $1 \times 62 = 62$

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Case 2: $a = 4$ and $b = 3$

Here, $|A \cap B| = \frac{4 \times 3}{6} = 2$.

• Number of ways to choose $A$: $\binom{6}{4} = 15$.
• To choose $B$, we select $2$ elements from $A$ and $1$ element from $S \setminus A$: $\binom{4}{2} \times \binom{6 - 4}{1} = \binom{4}{2} \times \binom{2}{1} = 6 \times 2 = 12$
• Total pairs for this case: $15 \times 12 = 180$

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Case 3: $a = 3$ and $b = 2$

Here, $|A \cap B| = \frac{3 \times 2}{6} = 1$.

• Number of ways to choose $A$: $\binom{6}{3} = 20$.
• To choose $B$, we select $1$ element from $A$ and $1$ element from $S \setminus A$: $\binom{3}{1} \times \binom{6 - 3}{1} = \binom{3}{1} \times \binom{3}{1} = 3 \times 3 = 9$
• Total pairs for this case: $20 \times 9 = 180$

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Total Number of Ordered Pairs:

$\text{Total} = 62 + 180 + 180 = 422$