Question

Tickets numbered from 1 to 18 are mixed up together and then a ticket is drawn at random. Find the probability that the ticket has a number, which is a multiple of 2 or 3.

• A. $\frac{1}{3}$
• B. $\frac{3}{5}$
• C. $\frac{2}{3}$
• D. $\frac{5}{6}$

Answer 1

The correct option is C $\frac{2}{3}$

$S=\{1,2,3,4,\cdots 18\}\Rightarrow n\left(S\right)=18E_1=\{2,4,6,8,10,12,14,16,18\}\Rightarrow n\left(E_1\right)=9E_2=\{3,6,9,12,15,18\}\Rightarrow n\left(E_2\right)=6(E_1\cap E_2)=E_3=\{6,12,18\}\Rightarrow n\left(E_3\right)=3\therefore E=E_1\cup E_2=E_1+E_2-E_{3}n\left(E\right)=9+6-3n\left(E\right)=12whereE=\{2,3,4,6,8,9,10,12,14,15,16,18\}\therefore P\left(E\right)=\frac{n\left(E\right)}{n\left(S\right)}=\frac{12}{18}=\frac{2}{3}$