If three distinct numbers are chosen randomly from first 100 natural numbers, then the probability that all…
If three distinct numbers are chosen randomly from first 100 natural numbers, then the probability that all three of them are divisible by both 2 and 3 is
$\frac{4}{35}$
$\frac{4}{55}$
$\frac{4}{1155}$
$\frac{80}{231}$
Solution
First 100 natural numbers are $\{1,2,3,4,5, \ldots . ., 100\}$
Numbers divisible by both 2 and 3 are $\{6,12,18, \ldots . ., 96\}$ (total 16)
Now the required probability $=\frac{{ }^{16} C_3}{{ }^{100} C_3}=\frac{\frac{\lfloor 16}{\left\lfloor\left\lfloor\frac{16-3}{\lfloor 100}\right.\right.}}{\frac{13\lfloor 100-3}{\lfloor\lfloor}}$
$\begin{aligned} & =\frac{\lfloor 16}{3 \underline{3} 13} \times \frac{\underline{3} \mid 97}{\underline{100}} \\ & =\frac{14 \times 15 \times 16}{98 \times 99 \times 100}=\frac{4}{1155} \\ & \end{aligned}$