A card is drawn from a pack of 52 cards. Find the probability that the card drawn is :

(i) a red card

(ii) a black card

(iii) a spade

(iv) an ace

(v) a black ace

(vi) ace of diamonds

(vii) not a club

(viii) a queen or a jack.

In a deck of 52 cards, there are 4 suits of 13 cards each.

No. of possible outcomes = 52.

(i) There are 26 red cards (13 hearts + 13 diamonds).

∴ No. of favourable outcomes = 26.

P(drawing a red card) = $\dfrac{\text{No. of favourable outcomes}}{\text{No. of possible outcomes}} = \dfrac{26}{52} = \dfrac{1}{2}$.

Hence, probability of drawing a red card = $\dfrac{1}{2}$.

(ii) There are 26 black cards (13 clubs + 13 spades).

∴ No. of favourable outcomes = 26.

P(drawing a black card) = $\dfrac{\text{No. of favourable outcomes}}{\text{No. of possible outcomes}} = \dfrac{26}{52} = \dfrac{1}{2}$.

Hence, probability of drawing a black card = $\dfrac{1}{2}$.

(iii) There are 13 spades in a deck of playing cards.

∴ No. of favourable outcomes = 13.

P(drawing a spade) = $\dfrac{\text{No. of favourable outcomes}}{\text{No. of possible outcomes}} = \dfrac{13}{52} = \dfrac{1}{4}$.

Hence, probability of drawing a spade = $\dfrac{1}{4}$.

(iv) There are 4 ace in a deck of playing cards.

∴ No. of favourable outcomes = 4.

P(drawing an ace) = $\dfrac{\text{No. of favourable outcomes}}{\text{No. of possible outcomes}} = \dfrac{4}{52} = \dfrac{1}{13}$.

Hence, probability of drawing an ace = $\dfrac{1}{13}$.

(v) There are 2 black ace (1 of club + 1 of spade).

∴ No. of favourable outcomes = 2.

P(drawing a black ace) = $\dfrac{\text{No. of favourable outcomes}}{\text{No. of possible outcomes}} = \dfrac{2}{52} = \dfrac{1}{26}$.

Hence, probability of drawing a black ace = $\dfrac{1}{26}$.

(vi) There is 1 ace of diamond.

∴ No. of favourable outcomes = 1.

P(drawing an ace of diamond) = $\dfrac{\text{No. of favourable outcomes}}{\text{No. of possible outcomes}} = \dfrac{1}{52}$.

Hence, probability of drawing an ace of diamond = $\dfrac{1}{52}$.

(vii) There are 13 clubs so, there are 39 (52 - 13) non-club cards.

∴ No. of favourable outcomes = 39.

P(drawing a non-club card) = $\dfrac{\text{No. of favourable outcomes}}{\text{No. of possible outcomes}} = \dfrac{39}{52} = \dfrac{3}{4}$.

Hence, probability of not drawing a club = $\dfrac{3}{4}$.

(viii) There are 4 queens and 4 jacks.

∴ No. of favourable outcomes = 8.

P(drawing a queen or a jack) = $\dfrac{\text{No. of favourable outcomes}}{\text{No. of possible outcomes}} = \dfrac{8}{52} = \dfrac{2}{13}$.

Hence, probability of drawing a queen or a jack = $\dfrac{2}{13}$.