Q. If three natural numbers from 1 to 100 are selected randomly, then probability that all are divisible by both 2 and 3, is
A) $\frac{4}{105}$
B) $\frac{4}{33}$
C) $\frac{4}{35}$
D) $\frac{4}{1155}$
Explanation
(d)
Let $E = E =$Events of numbers divisible by 2 and 3 [$ie$, divisible by 6]
$= (6,\ 12,\ldots,96)$
$n(E) = 16$
$\therefore$Required probability $= \frac{\ ^{16}C_{3}}{\ ^{100}C_{3}}$
$= \frac{\frac{16 \times 15 \times 14}{3 \times 2 \times 1}}{\frac{100 \times 99 \times 98}{3 \times 2 \times 1}} = \frac{4}{1155}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.