Q. Twelve tickets are numbered from 1 to 12. One ticket is drawn at random, then the probability of the number to be divisible by 2 or 3, is
A) $\frac{2}{3}$
B) $\frac{7}{12}$
C) $\frac{5}{6}$
D) $\frac{3}{4}$
Explanation
(a)
Let $E$ be the event of numbers to be divisible by 2 or 3
$\therefore\ E = \left\{ 2,\ 3,\ 4,\ 6,\ 8,\ 9,\ 10,12 \right\}$
$\Rightarrow \ n(E) = 8\$and $n(S) = 12$
$Hence,\ required\ probability = \frac{n(E)}{n(S)}$
$= \frac{8}{12} = \frac{2}{3}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.