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Question
Four cards are drawn at random from a well-shuffled pack of 52 cards. Find the probability that there is exactly one pair.
Four cards are drawn at random from a well-shuffled pack of 52 cards. Find the probability that there is exactly one pair.
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Solution
The correct option is A (13C1×4C2×48C1×44C1)52C4
Let H, C, D, S denotes heart, club, diamond and spade respectively.WE have a pack of cards like this:1H, 1C, 1D ,1S(Aces) .........(1)2H, 2C, 2D, 2S ........(2)....JH, JC, JD, JS .......(11)QH, QC, QD, QS .......(12)KH, KC, KD, KS .......(13)We now select a pair. This can be done by firstly selecting a row and then from the selected row we select a pair out of it.There are 13 rows and a row is selected in 13C1 ways. Each row contain 4 elements and a pair from the selected row is selected in 4C2 ways.Number of ways a pair is selected selected 13C1×4C2 ways.Remaining two cards has to be selected so that is does not pair with any of other three cards. So the third card has to be selected from those cards which do not belong to the row we selected earlier, i.e., selection has to be made from remaining 48 cards. Third card can be selected in 48C1 ways.For the fourth card we do not select that card which belongs to either the row we selected or the row to which third card belongs. So selection of fourth card is made from remaining 44 cards in 44C1 ways.Therefore, number of ways of selecting four cards = 13C1×4C2×48C1×44C1Total number number of ways = 52C4Therefore required probability = 13C1×4C2×48C1×44C152C4
Let H, C, D, S denotes heart, club, diamond and spade respectively.
WE have a pack of cards like this:
1H, 1C, 1D ,1S(Aces) .........(1)
2H, 2C, 2D, 2S ........(2)
....
JH, JC, JD, JS .......(11)
QH, QC, QD, QS .......(12)
KH, KC, KD, KS .......(13)
We now select a pair. This can be done by firstly selecting a row and then from the selected row we select a pair out of it.
There are 13 rows and a row is selected in 13C1 ways. Each row contain 4 elements and a pair from the selected row is selected in 4C2 ways.
Number of ways a pair is selected selected 13C1×4C2 ways.
Remaining two cards has to be selected so that is does not pair with any of other three cards. So the third card has to be selected from those cards which do not belong to the row we selected earlier, i.e., selection has to be made from remaining 48 cards. Third card can be selected in 48C1 ways.
For the fourth card we do not select that card which belongs to either the row we selected or the row to which third card belongs. So selection of fourth card is made from remaining 44 cards in 44C1 ways.
Therefore, number of ways of selecting four cards = 13C1×4C2×48C1×44C1
Total number number of ways = 52C4
Therefore required probability = 13C1×4C2×48C1×44C152C4
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