In a poker hand consisting of 5 cards, find the probability of holding (a) 2 aces; (b) 4 spades and 1 red card

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In a poker hand consisting of 5 cards, find the probability of holding (a) 2 aces; (b) 4 spades and 1 red card&amp;nbsp;

karton · Accepted Answer

To find the probabilities of holding certain combinations of cards in a poker hand consisting of 5 cards, we need to consider the total number of possible hands and the number of favorable outcomes for each case.
(a) Probability of holding 2 aces:
In a standard deck of 52 cards, there are 4 aces. To calculate the probability of holding 2 aces, we need to consider the number of ways to choose 2 aces out of 4, and the remaining 3 cards can be any of the remaining 48 cards.
The total number of possible hands is the number of ways to choose any 5 cards out of 52, which can be calculated using combinations:

(\binom{52}{5})

.
The number of favorable outcomes is the number of ways to choose 2 aces out of 4 and the remaining 3 cards out of the remaining 48, which can be calculated as:

(\binom{4}{2}) \cdot (\binom{48}{3})

.
Therefore, the probability of holding 2 aces is given by:

P (2 aces) = \frac{(\binom{4}{2}) \cdot (\binom{48}{3})}{(\binom{52}{5})}

(b) Probability of holding 4 spades and 1 red card:
In a standard deck of 52 cards, there are 13 spades and 39 red cards (clubs and diamonds). To calculate the probability of holding 4 spades and 1 red card, we need to consider the number of ways to choose 4 spades out of 13 and 1 red card out of 39.
The total number of possible hands is still

(\binom{52}{5})

.
The number of favorable outcomes is the number of ways to choose 4 spades out of 13 and 1 red card out of 39, which can be calculated as:

(\binom{13}{4}) \cdot (\binom{39}{1})

.
Therefore, the probability of holding 4 spades and 1 red card is given by:

P (4 spades and 1 red card) = \frac{(\binom{13}{4}) \cdot (\binom{39}{1})}{(\binom{52}{5})}

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In a poker hand consisting of 5 cards,

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