Question

Let $|X|$ denote title 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 5, then the number of ordered pairs (A, B) such that $1\le \left|B\right|<\left|A\right|,$ equals

Answer 1

Answer: 422

$P\left(\frac{B}{A}\right)=P\left(B\right)$
$\Rightarrow n\frac{\left(A\cap B\right)}{n\left(A\right)}=\frac{n\left(B\right)}{n\left(S\right)}........\left(1\right)$
$\Rightarrow$ $n\left(A\right)$ should have 2 or 3 as prime factors
$\Rightarrow n\left(A\right)$ can be $2,3,4$ or $6$ as $n\left(A\right)>1$
$n\left(A\right)=2$ does not satisfy the constraint $\left(1\right)$.
for $n\left(A\right)=3$. $n\left(B\right)=2$ and $n\left(A\cap B\right)=1$
$\Rightarrow$ No. of ordered pair ${=}^6{C}_4\times \frac{4!}{2!}=180$
for $n\left(A\right)=4.n\left(B\right)=3$ and $n\left(A\cap B\right)=2$
$\Rightarrow$ No. of ordered pairs ${=}^6{C}_5\times \frac{5!}{2!2!}=180$
for $n\left(A\right)=6.n\left(B\right)$ can be $1,2,3,4,5.$
$\Rightarrow$ No. of ordered pairs $=2^6-2=62$
Total ordered pair $=180+180+62=422.$