Question

If a,b and c are distinct natural numbers and a,b,c > $100$, then probability that a,b and c are individually divisible by both $2$ and $3$ is,

• A. $\frac{5}{28}$
• B. $\frac{4}{1155}$
• C. $\frac{3}{28}$
• D. $\frac{5}{29}$

Answer 1

The correct option is B $\frac{4}{1155}$

A natural number is divisible by $2$ or $3$ if it is divisible by $6$.

So, numbers divisible by 6 are $6,12,18,....,96$

So there are total $16$ numbers out of which $3$ distinct number can be chosen in $16\times 15\times 14=n\left(A\right)$

Total no of ways $=100\times 99\times 98=n\left(S\right)$

Required probability= $\frac{n\left(A\right)}{n\left(S\right)}=\frac{4}{1155}.$

Hence option $B$ is the answer.