Q260 / 270 · NDA 2024 (Phase 1) · Mathematics

Three different numbers are selected at random from the first 15 natural numbers. What is the probability that the product of two of the numbers is equal to third number?

  • A
    $\frac{1}{91}$
  • B
    $\frac{2}{455}$
  • C
    $\frac{1}{65}$
  • Correct answer
    D
    $\frac{6}{455}$

Solution

$\begin{aligned} n(\mathrm{~S}) & ={ }^{15} \mathrm{C}_3 \\ 2 \times 3 & =6 \\ 2 \times 4 & =8 \\ 2 \times 5 & =10 \\ 2 \times 6 & =12 \\ 2 \times 7 & =14 \\ 3 \times 4 & =12\end{aligned}$ $\therefore \quad n(\mathrm{E})=6$ $\therefore$ Required probability $=\frac{6}{{ }^{15} \mathrm{C}_3}=\frac{6}{\frac{15 \times 14 \times 13 \times 12!}{12!\times 3 \times 2}}$ $=\frac{6}{35 \times 13}=\frac{6}{455}$

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

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