Question

Let $|X|$ denotes 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 $1\le |B|<|A|,$ equals

Answer 1

As, $A$ and $B$ are independent events.
So, $P\left(\frac{B}{A}\right)=P\left(B\right)$
$P(A\cap B)=P\left(A\right)\cdot P\left(B\right)$
Let number of elements in $A=a$ and number of elements in $B=b[\because a>b\ge 1]$
and number of elements in $A\cap B=z.$
$P\left(A\right)=\frac{n\left(A\right)}{n\left(S\right)}=\frac{a}{6}$
$P\left(B\right)=\frac{n\left(B\right)}{n\left(S\right)}=\frac{b}{6}$
$P(A\cap B)=\frac{n(A\cap B)}{n\left(S\right)}=\frac{z}{6}$
$\frac{z}{6}=\frac{a}{6}\cdot \frac{b}{6}\Rightarrow 6z=a\cdot b$

Now case 1: if $z=1$
$a=6,b=1\Rightarrow {}^6{C}_6\times {}^6{C}_1=6$
$a=3,b=2\Rightarrow {}^6{C}_3\times {}^3{C}_1\times {}^3{C}_1=180$
case 2: if $z=2$
$a=6,b=2\Rightarrow {}^6{C}_6\times {}^6{C}_2=15$
$a=4,b=3\Rightarrow {}^6{C}_4\times {}^4{C}_2\times {}^2{C}_1=180$
case 3: if $z=3$
$a=6,b=3\Rightarrow {}^6{C}_6\times {}^6{C}_3=20$
case 4: if $z=4$
$a=6,b=4\Rightarrow {}^6{C}_6\times {}^6{C}_4=15$
case 5: if $z=5$
$a=6,b=5\Rightarrow {}^6{C}_6\times {}^6{C}_5=6$
Hence number of ordered pairs $(A,B)=6+180+20+15+180+15+6=422.$