Question

If three distinct numbers are chosen randomly from the first $100$ natural numbers, then the probability that all the three numbers are divisible by both $2$ and $3$ is

• A. $\frac{4}{57}$
• B. $\frac{7}{99}$
• C. $\frac{4}{1155}$
• D. $\frac{1}{1100}$

Answer 1

Answer: (C)

$\frac{4}{1155}$

The numbers are divisible by both 2 and 3 means no.'s divisible by 6.

No. of no.'s divisible by $6$ in $1$ to $100$ are $16$

No. of ways of selecting $3$ of them are ${16}_{C_3}$

Probability $=\frac{{16}_{C_3}}{{100}_{C_3}}=\frac{16\times 15\times 14}{100\times 99\times 98}$

$=\frac{4}{1155}$