Q. From a set of 100 cards numbered 1 to 100, one card is drawn at random. The probability that the number obtained on the card is divisible by 6 or 8 but not by 24, is
A) $\frac{6}{25}$
B) $\frac{1}{4}$
C) $\frac{1}{6}$
D) $\frac{1}{5}$
Explanation
(d)
$E =$Set of number divisible by 6
$= \{ 6,12,18,24,30,\ldots..,96\}$
$F =$Set of number divisible by 8
$= \left\{ 8,16,24,\ldots.,96 \right\}$
$E \cap F =$Set of number divisible by 24
$= \{ 24,48,72,96\}$
$\therefore n(E) = 16,\ n(F) = 12.n(E \cap F) = 4$
$Let\ n(A) = n(E) + n(F) - 2n(E \cap F)$
$= 16 + 12 - 8 = 20$
and $n(S) = 100$
$\therefore$ Required probability
$= \frac{n(E \cup F)}{n(S)} = \frac{20}{100} = \frac{1}{5}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.