Question

Find the probability that when a hand of 7 cards are drawn from the well-shuffled deck of 52 cards, it contains

(i) all kings (ii) 3 kings

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Answer 1

Answer:

Total number of possible hands = ⁵²C₇

$\tt \red {i) Number \: \: of \: \: hands \: \: with \: \: 4 \: kings = \: ⁴C₄ \: × \: ⁴⁸C₃}$

(other 3 cards must be chosen from the rest 48 cards )

Hence, P ( a hand will have 4 Kings )

${\tt{ \green{ \pink{ = ⁴C₄ × ⁴⁸C₃ / ⁵²C₇ = \frac{1}{7735} }}}}$

$\red{ \tt{ii) Number \: \: of \: \: hands \: \: with \: \: 3 \: \: Kings \: \: and \: \: 4 \: \: non-Kings \: \: cards \: \: = ⁴C₃ × ⁴⁸C₄ / ⁵²C₇}}$

$\red{= \frac{9}{1547} }$

Step-by-step explanation:

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Answer 2

Total number of possible hands = ⁵²C₇

i) Number of hands with 4 kings QA

• = ⁴C₄ × ⁴⁸C₃

(other 3 cards must be chosen from the rest 48 cards )

Hence, P ( a hand will have 4 Kings )

$= ⁴C₄ × ⁴⁸C₃ / ⁵²C₇ = \frac{1}{7735}$

ii) Number of hands with 3 Kings and 4 non-Kings cards

$= > ⁴C₃ × ⁴⁸C₄ / ⁵²C₇ = \frac{9}{1547}$

★Hope it helps you.