Q. One ticket is selected at random from 50 tickets numbered 00,01,02,….,49. Then, the probability that the sum of the digits on the selected ticket is 8, given that the product of these digits is zero equals
A) $\frac{1}{14}$
B) $\frac{1}{7}$
C) $\frac{5}{14}$
D) $\frac{1}{50}$
Explanation
(a)
$S = \{ 00,01,\ 02,\ldots,49\}$
Let $A$ be the event that sum of the digits on the selected ticket is 8, then
$A = \{ 08,\ 17,\ 26,\ 35,\ 44\}$
Let $B$be the event that the product of the digits is zero
$B =${00, 01, 02, 03,….,09, 10, 20, 30, 40}
$\therefore A \cap B = \{ 8\}$
$\therefore$Required probability=$P\left( \frac{A}{B} \right)$
$= \frac{P(A \cap B)}{P(B)} = \frac{\frac{1}{50}}{\frac{14}{50}} = \frac{1}{14}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.