Search any question & find its solution
Question:
Answered & Verified by Expert
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
Options:
Solution:
2726 Upvotes
Verified Answer
The correct answer is:
$\frac{4}{1155}$
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 $=\frac{{ }^{16} C_3}{{ }^{100} C_3}=\frac{\frac{\lfloor 16}{\left\lfloor\left\lfloor\frac{16-3}{\lfloor 100}\right.\right.}}{\frac{13\lfloor 100-3}{\lfloor\lfloor}}$
$\begin{aligned} & =\frac{\lfloor 16}{3 \underline{3} 13} \times \frac{\underline{3} \mid 97}{\underline{100}} \\ & =\frac{14 \times 15 \times 16}{98 \times 99 \times 100}=\frac{4}{1155} \\ & \end{aligned}$
Numbers divisible by both 2 and 3 are $\{6,12,18, \ldots . ., 96\}$ (total 16)
Now the required probability $=\frac{{ }^{16} C_3}{{ }^{100} C_3}=\frac{\frac{\lfloor 16}{\left\lfloor\left\lfloor\frac{16-3}{\lfloor 100}\right.\right.}}{\frac{13\lfloor 100-3}{\lfloor\lfloor}}$
$\begin{aligned} & =\frac{\lfloor 16}{3 \underline{3} 13} \times \frac{\underline{3} \mid 97}{\underline{100}} \\ & =\frac{14 \times 15 \times 16}{98 \times 99 \times 100}=\frac{4}{1155} \\ & \end{aligned}$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.