Question

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

• A. $\frac{4}{25}$
• B. $\frac{4}{35}$
• C. $\frac{4}{1161}$
• D. $\frac{4}{1155}$

Answer 1

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

A number is divisible by both $2$ and $3$ if it is divisible by $6$,
The numbers that are divisible by $6$ are
$\{6,12,18,24,,,,,96\}=16$
Required probabilty
$=\frac{{}^{16}{C}_3}{{}^{100}{C}_3}=\frac{16\times 15\times 14}{100\times 99\times 98}=\frac{4}{1155}$