From a pack of 52 cards,4 cards are drawn. Wh | vedclass

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(C) Total number of ways to draw $4$ cards from $52$ is $n(S) = {}^{52}C_4 = \frac{52 	imes 51 	imes 50 	imes 49}{4 	imes 3 	imes 2 	imes 1} = 2

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$\frac{1152}{7735}$

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(C) Total number of ways to draw $4$ cards from $52$ is $n(S) = {}^{52}C_4 = \frac{52 	imes 51 	imes 50 	imes 49}{4 	imes 3 	imes 2 	imes 1} = 270725$.Digit cards are cards numbered $2$ to $10$,so there are $9$ cards per suit,total $36$ digit cards.Honours cards are Ace,King,Queen,Jack,so there are $4$ cards per suit.We need $2$ digit cards,$1$ black honours card,and $1$ red honours card.Number of ways to choose $2$ digit cards from $36$ is ${}^{36}C_2 = \frac{36 	imes 35}{2} = 630$.Number of ways to choose $1$ black honours card from $8$ (spades and clubs) is ${}^{8}C_1 = 8$.Number of ways to choose $1$ red honours card from $8$ (hearts and diamonds) is ${}^{8}C_1 = 8$.Total favorable outcomes $n(E) = 630 	imes 8 	imes 8 = 40320$.Probability $P(E) = \frac{40320}{270725} = \frac{1152}{7735}$.

From a pack of $52$ cards,$4$ cards are drawn. What is the probability that it has two digit-cards and one honours card of black suit and one honours card of red suit?

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