Question

If three natural numbers from $1$ to $100$ are selected randomly then probability that all are divisible by both $2$ and $3$, is

• A. $\frac{4}{105}$
• B. $\frac{4}{33}$
• C. $\frac{4}{35}$
• D. $\frac{4}{1155}$

Answer 1

Answer: (D)

If a number is divisible by both $2$ and $3$ that means it is divisible by $6$.
So sample space $S$ is a number divisible by $6$, is
$S=\{6,12,18,24,30,36,42,48,54,60,66,72,78,84,90,96\}$
$\Rightarrow n(S)=16$
So, required probability $\frac{{}^{16}{C}_3}{{}^{100}{C}_3}$
$=\frac{16\times 15\times 14}{100\times 99\times 98}$
$=\frac{4}{1155}$