Question

Suppose you draw 5 cards from a standard 52 card deck (13 rank cards in 4 suits). What is the probability your hand contains at least two aces or at least two kings?

Answer 1

Solution:

Suppose you draw 5 cards from the standard 52 card deck,

The probability that hand contains at least two aces or at least kings.

P(at least two aces or at least two kings)

= P (at least two aces) + P (at least two kings) - P (at least two aces $\cap$ at least two kings)

Now,

P (at least two aces) = P (2 aces, 3 others) + P (3 aces, 2 others), + P (4 aces, 1 other)

(3) (a) (23) (5) ()() + (S)( )+(5)

P (at least two kings) = P (2 kings, 3 others) + P (3 kings, 2 others), + P (4 kings, 1 other)

(3) (a) (23) (5) ()() + (S)( )+(5)

P (at least two aces $\cap$ at least two kings) = P (2 aces, 3 kings) + P (3 aces, 2 kings) + P (2 aces, 2 kings, 1 other)

$=\frac{ \binom{4}{2}*\binom{4}{3}}{\binom{52}{5}}+\frac{ \binom{4}{3}*\binom{4}{2}}{\binom{52}{5}}+\frac{ \binom{4}{2}*\binom{4}{2}*\binom{44}{1}}{\binom{52}{5}}$

Therefore,

P (at least two aces or at least two kings)

= P (at least two aces) + P (at least two kings) - P (at least two aces $\cap$ at least two kings)

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