Two cards are drawn from a well shuffled pace of 52 cards. Find the pr

1.	Two cards are drawn from a well shuffled pace of 52 cards. Find the probability that(1) both the cards are of different colour(2) both the cards are face cards(3) one of the two cards is a king.
Answer» Total number of primary outcomes of drawing two cards from a well shuffled pack of 52 cards is n = ⁵²C₂ = \(\frac{52×51}{2×1} =1326\) (1) A = Event that both the cards drawn are of different colours, i.e., one card is of black colour and one is of red colour. In a pack of 52 cards, 26 cards are black and 26 cards are red. ∴ Favourable outcomes for the event A is m = ²⁶C₁ × ²⁶C₁ = 26 × 26 = 676 Hence, P(A) = \(\frac{m}{n} = \frac{676}{1326} = \frac{26}{51}\) (2) B = Event that both the cards are face cards. In a pack of 52 cards, 12 cards are face cards. ∴ Favourable outcomes for the event B is m = ¹²C₂ = \(\frac{12×11}{2×1} = 66.\) Hence, p(B) = \(\frac{m}{n} = \frac{66}{1326} = \frac{11}{221}\) (3) C = Event that one of the two cards is a king. In a pack of 52 cards, 4 cards are of king and other cards are 48. ∴ Favourable outcomes for the event C is m = ⁴C₁ × ⁴⁸C₁ = 4 × 48 = 192 Hence, P(C) =\(\frac{ m}{n} = \frac{192}{1326} = \frac{32}{221}\)

Two cards are drawn from a well shuffled pace of 52 cards. Find the probability that(1) both the cards are of different colour(2) both the cards are face cards(3) one of the two cards is a king.

Answer»

Total number of primary outcomes of drawing two cards from a well shuffled pack of 52 cards is

n = ⁵²C₂ = \(\frac{52×51}{2×1} =1326\)

(1) A = Event that both the cards drawn are of different colours, i.e., one card is of black colour and one is of red colour.

In a pack of 52 cards, 26 cards are black and 26 cards are red.

∴ Favourable outcomes for the event A is

m = ²⁶C₁ × ²⁶C₁ = 26 × 26 = 676

Hence, P(A) = \(\frac{m}{n} = \frac{676}{1326} = \frac{26}{51}\)

(2) B = Event that both the cards are face cards. In a pack of 52 cards, 12 cards are face cards.

∴ Favourable outcomes for the event B is

m = ¹²C₂ = \(\frac{12×11}{2×1} = 66.\)

Hence, p(B) = \(\frac{m}{n} = \frac{66}{1326} = \frac{11}{221}\)

(3) C = Event that one of the two cards is a king. In a pack of 52 cards, 4 cards are of king and other cards are 48.

∴ Favourable outcomes for the event C is

m = ⁴C₁ × ⁴⁸C₁ = 4 × 48 = 192

Hence, P(C) =\(\frac{ m}{n} = \frac{192}{1326} = \frac{32}{221}\)