Question

Tickets numbered from 1 to 12 are mixed up together and then a ticket is withdrawn at random. Find the probability that the ticket has a number which is a multiple of 2 or 3.

Answer 1

Clearly, n(S) = 12 and $E=2,4,6,8,10,12,3,9\Rightarrow n\left(E\right)=8$

$\therefore P\left(E\right)=\frac{n\left(E\right)}{n\left(S\right)}=\frac{8}{12}=\frac{2}{3}$