If the probability that a randomly chosen 6 -digit number formed by...

We have to choose $6$ digit number by using $1 \text{ and } 8$ and it must be divisible by 21,

So, equal number of digits $1$ and $8$ should be used to form a number divisible by $3$ and $7$. (Or all the digits are either $1$ or $8$)

In all cases whatever number is formed using six ${1}^{'}$ or six ${8}^{'}$s or three ${1}^{'}$s $+$ three ${8}^{'}$s, that number can be splited into sum of multiples of $1001, 1008, 8001, 8008$ (all of them are divisible by $7$). So, all the numbers thus formed will be divisible by $21$,

Example: $811188 = 8001 \times {10}^{2} + 1008 \times 10 + 1008$

Number of such numbers $= \dfrac{6 !}{3 ! 3 !} + 2 = 22$

Required Probability, $p = \dfrac{22}{{2}^{6}} = \dfrac{11}{32}$

$∴ 96p = \dfrac{11}{22} \times 96 = 33$