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

$P\left(\frac{B}{A}\right)=P\left(B\right)$
$\Rightarrow n \frac{\left(A\cap B\right)}{n\left(A\right)}=\frac{n\left(B\right)}{n\left(S\right)} ........\left(1\right)$
$\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 $= 180 + 180 + 62 = 422.$