If the probability that a randomly chosen 6 -digit number formed by using digits 1 and 8 only is a multiple…

We have to choose $6$ digit number by using $1 \& 8$ 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 $1^{\text{'}}$ or six $8^{\text{'}}$s or three $1^{\text{'}}$s $+$ three $8^{\text{'}}$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 $=\frac{6!}{3!3!}+2=22$

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

$\therefore 96p=\frac{11}{22}\times 96=33$