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)

$\frac{4}{1155}$

Answer 2

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 $\frac{^{16}C_{3}}{^{100}C_{3}}$
$=\frac{ 16\times15\times14}{100\times99\times98}$
$= \frac{4}{1155}$