MHT CET · Maths · Probability
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
- A \(\frac{4}{35}\)
- B \(\frac{4}{55}\)
- C \(\frac{4}{1155}\)
- D \(\frac{80}{231}\)
Answer & Solution
Correct Answer
(C) \(\frac{4}{1155}\)
Step-by-step Solution
Detailed explanation
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 \(=\frac{{ }^{16} C_3}{{ }^{100} C_3}=\frac{\frac{\lfloor 16}{\left\lfloor\left\lfloor\frac{16-3}{\lfloor 100}\right.\right.}}{\frac{13\lfloor 100-3}{\lfloor\lfloor}}\)
\(\begin{aligned} & =\frac{\lfloor 16}{3 \underline{3} 13} \times \frac{\underline{3} \mid 97}{\underline{100}} \\ & =\frac{14 \times 15 \times 16}{98 \times 99 \times 100}=\frac{4}{1155} \\ & \end{aligned}\)
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