Question

From a set of 100 cards numbered 1 to 100, one card is drawn at random. Find the probability that the number obtained on the card is divisible by 6 or 8, but not by 24.

• (A) $\frac{6}{25}$
• (B) $\frac{1}{4}$
• (C) $\frac{1}{6}$
• (D) $\frac{2}{5}$

Answer 1

The Correct Option is A

Explanation:
$E=\mathrm{set}$ of numbers divisible by 6$$\begin{array}{r}=\{6,12,18,24,30,\dots ,96\}\\ F=\mathrm{set}\text{ of numbers divisible by }8\\ =\{8,16,24,\dots ,96\}\end{array}$$$E\cap F=\mathrm{set}$ of numbers divisible by 6 and 8$$\begin{array}{l}\begin{array}{r}=\{24,48,72,96\}\\ \therefore n(E)=16,& n(F)=12,n(E\cap F)=4\\ \therefore n(E\cup F)=n(E)+n(F)-n(E\cap F)\\ =& 16+12-4=24\\ \text{ and }& n(S)=100\end{array}\end{array}$$$\therefore$ Required probability$$=\frac{n(E\cup F)}{n(S)}=\frac{24}{100}=\frac{6}{25}$$