# Number of ordered pairs of $A,B$ in Probability

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

what i try

Let number of elements in $$A,B,A \cap B$$ is $$x,y,z$$ respectively

conditions given as $$1\leq y\leq x.$$

For $$2$$ Independent events $$A,B$$

is $$\displaystyle P(A \cap B)=P(A)\cdot P(B)\Rightarrow \frac{z}{6}=\frac{x}{6}\cdot \frac{y}{6}\Rightarrow z=xy/6$$

Given condition is |A|>|B| , A and B are independent events which gives total number of ordered pairs (A,B) = 422

The equation $$z=\frac{xy}{6}$$ shows that the events $$A$$ and $$B$$ can be independent only if

1. $$A=S$$, or
2. $$|A|=3,|B|=2$$
3. $$|A|=4,|B|=3$$

In the 2nd case $$A$$ and $$B$$ have 1 common element; in the 3rd case $$A$$ and $$B$$ have 2 common elements.

In the 1st case, we have $$2^6-2$$ solutions ($$B$$ can be anything except $$S$$ and $$\varnothing$$).

In the 2nd case, we can choose $$A$$ by $$\binom{6}{3}$$ ways and then $$B$$ by $$3\cdot 3$$ ways.

In the 3rd case we can choose common elements by $$\binom{6}{2}$$ ways, then 2 remaining A elements by $$\binom{4}{2}$$ ways and the remaining B element by 2 ways.

Summing it all $$64-2+\frac{6\cdot 5\cdot 4}{6}\cdot 3 \cdot 3 + \frac{6\cdot 5}{2}\cdot \frac{4\cdot 3}{2}\cdot 2= 422$$

PS: thanks @karthikeya kurella for noticing the 3rd case.

The condition should be $$1\le y\lt x$$, not $$\le$$.

Since $$z=xy/6$$, $$x$$ or $$y$$ must be divisible by $$2$$ and by $$3$$, and $$xy\le6$$. With $$1\le y\lt x$$, that allows only two solutions: $$x=6$$, $$y=1$$ and $$x=3$$, $$y=2$$.

In the first case, we have $$6$$ choices for the element in $$y$$.

In the second case we have $$z=3\cdot2/6=1$$, so there must be exactly one common element. That gives us $$6$$ choices for the common element, then $$5$$ choices for the other element in $$y$$ and then $$\binom42=6$$ choices for the other two elements in $$x$$.

Thus the total number of ordered pairs is $$6+6\cdot5\cdot6=186$$.