Question

from a set of 12 cards numbered 1,2,3........12,one is drawn. Find the probability of getting a multiple of 3

Answer 1

`|\Omega|=12\\|A|=4\\\\P(A)=(4)/(12)=(1)/(3)\approx33\%`

Answer 2

These exercise are solved by counting how many outcomes are in your favour, over how many outcomes are there in total.

Obviously, there are 12 possible outcomes, because you may pick any number from 1 to 12.

The favourable cases are the multuples of 3 which lie between 1 and 12, which are 3, 6, 9, 12.

So, there are 4 favourable cases over 12 possible cases, which leads to a probability of

`(4)/(12) = (1)/(3)`