One card is drawn from a pack of 52 cards, each of the 52 cards being equally likely to be drawn. Find the probability that the card drawn is:

- (i) an ace
- (ii) red
- (iii) either red or king
- (iv) red and a king

**Solution**

Out of 52 cards, one card can be drawn in ^{52}C_{1} ways.

Therefore exhaustive number of cases = ^{52}C_{1} = 52.

(i) There are four aces in a pack of 52 cards, out of which one can be drawn in ^{4}C_{1} ways.

Therefore, favourable number of cases = ^{4}C_{1} = 4

So, required probability = 4/52 = 1/13

(ii) There are 26 red cards, out of which one red card can be drawn in ^{26}C_{1} ways.

Therefore, favourable number of cases = ^{26}C_{1} = 26

So, required probability = 26/52 = 1/2

(iii) There are 26 red cards including 2 red kings and there are 2 more kings. Therefore, there are 28 cards out of which one can be drawn in ^{28}C_{1} ways.

Therefore, favourable number of cases = ^{28}C_{1} = 28.

So, required probability = 28/52 = 7/13

(iv) There are 2 cards, which are red and king. Therefore, favourable number of cases = ^{2}C_{1} = 2

So, required probability = 2/52 = 1/26