In a poker hand consisting of 5 cards,

Annabathuni Seethu keerthana

Annabathuni Seethu keerthana

Answered question

2022-07-24

In a poker hand consisting of 5 cards, find the probability of holding (a) 3 aces (b) 4 hearts and 1 club (c) Cards of same suit (d) 2 aces and 3 jacks

Answer & Explanation

karton

karton

Expert2023-06-02Added 613 answers

(a) Finding the probability of holding 3 aces:
To find the probability of holding 3 aces in a poker hand, we need to determine the total number of possible poker hands and the number of hands that contain 3 aces.
Total number of possible poker hands = (525) (since we choose 5 cards from a standard deck of 52 cards)
Number of hands with 3 aces = (43)·(482) (we choose 3 aces from the 4 available and 2 additional cards from the remaining 48 non-ace cards)
Therefore, the probability of holding 3 aces is given by:
P(3 aces)=(43)·(482)(525)
(b) Finding the probability of holding 4 hearts and 1 club:
To find the probability of holding 4 hearts and 1 club, we again need to determine the total number of possible poker hands and the number of hands that satisfy this condition.
Total number of possible poker hands = (525)
Number of hands with 4 hearts and 1 club = (134)·(131) (we choose 4 cards from the 13 available hearts and 1 card from the 13 available clubs)
Therefore, the probability of holding 4 hearts and 1 club is given by:
P(4 hearts and 1 club)=(134)·(131)(525)
(c) Finding the probability of holding cards of the same suit:
To find the probability of holding cards of the same suit, we need to determine the total number of possible poker hands and the number of hands where all 5 cards belong to the same suit.
Total number of possible poker hands = (525)
Number of hands with cards of the same suit = 4·(135) (we choose 5 cards from any of the 4 suits)
Therefore, the probability of holding cards of the same suit is given by:
P(cards of the same suit)=4·(135)(525)
(d) Finding the probability of holding 2 aces and 3 jacks:
To find the probability of holding 2 aces and 3 jacks, we once again need to determine the total number of possible poker hands and the number of hands that satisfy this condition.
Total number of possible poker hands = (525)
Number of hands with 2 aces and 3 jacks = (42)·(43) (we choose 2 aces from the 4 available and 3 jacks from the 4 available)
Therefore, the probability of holding 2 aces and 3 jacks is given by:
P(2 aces and 3 jacks)=(42)·(43)(525)

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