Q99 / 150 · MHT CET 2022 (05 Aug Shift 1) · Mathematics

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
    $\frac{4}{55}$
  • Correct answer
    C
    $\frac{4}{1155}$
  • D
    $\frac{80}{231}$

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 $=\frac{{ }^{16} C_3}{{ }^{100} C_3}=\frac{\frac{\lfloor 16}{\left\lfloor\left\lfloor\frac{16-3}{\lfloor 100}\right.\right.}}{\frac{13\lfloor 100-3}{\lfloor\lfloor}}$ $\begin{aligned} & =\frac{\lfloor 16}{3 \underline{3} 13} \times \frac{\underline{3} \mid 97}{\underline{100}} \\ & =\frac{14 \times 15 \times 16}{98 \times 99 \times 100}=\frac{4}{1155} \\ & \end{aligned}$

Hence, the correct answer is option: 'C'.

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