Q. Three numbers are chosen at random from 1 to 20. The probability that they are consecutive, is
A) $\frac{1}{190}$
B) $\frac{1}{120}$
C) $\frac{3}{190}$
D) $\frac{5}{190}$
Explanation
(c)
The total number of favourable cases, $n(E) = 18$
The total number of cases, $n(S) = \ ^{20}C_{3}$
$= \frac{20 \times 19 \times 18}{3 \times 2 \times 1} = 1140$
$\therefore Required\ probability = \frac{18}{1140} = \frac{3}{190}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.