Q103 / 120 · NDA 2024 (Phase 2) · Mathematics

Three distinct natural numbers are chosen at random from 1 to 10 . What is the probability that they are consecutive?

  • A
    $\frac{1}{12}$
  • B
    $\frac{3}{40}$
  • Correct answer
    C
    $\frac{1}{15}$
  • D
    $\frac{7}{120}$

Solution

Three distinct natural numbers are chosen from 1 to 10 . $\mathrm{n}(\mathrm{s})=10 \mathrm{c}_3$ Favourable outcomes $=(1,2,3),(2,3,4),(3,4,5),(4,5$, $6),(5,6,7),(6,7,8),(7,8,9),(8,9,10)$ $\mathrm{n}(\mathrm{E})=8$ $\therefore \quad P(E)=\frac{n(E)}{n(S)}=\frac{8}{10_3}=\frac{8 \times 7!\times 3!}{10!}=\frac{8 \times 3 \times 2 \times 1}{10 \times 9 \times 8}=\frac{1}{15}$

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

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