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In a pack of cards there are 52 cards: cards of 13 denominations belonging to each of 4 suits (clubs, diamonds, hearts and spades). 5 cards are drawn from a pack of 52 cards. What is the probability of getting 3 cards of one denomination and 2 cards of another denomination? the way i approached is [13C3+39C2/52C5 ]*4 is this a right way of doing ?

In a pack of cards there are 52 cards: cards of 13 denominations belonging to each of 4 suits (clubs, diamonds, hearts and spades). 5 cards are drawn from a pack of 52 cards. What is the probability of getting 3 cards of one denomination and 2 cards of another denomination?
 
the way i approached is [13C3+39C2/52C5 ]*4 
is this a right way of doing ?

Grade:12th pass

1 Answers

Rajat
213 Points
5 years ago
First you have to choose a denomination out of the 13 denominations, that can be done in 13C1 ways= 13. There are four cards present of the chosen enomination from the four suits, we have to choose 3 among them = 4C3= 4. 
We are left out with 12 different denominations (except the chosen one) we have to choose one denomination out of these 12 denominations.
No of ways = 12C1= 12, in the chosen denomination we have four cards from 4 different suits and have to choose 2 among them.
Ways = 4C2 = 6
the event of choosing 3 cards from denomination and 2 cards from a different denomination are independent of each other thus we have to multiply them.
Required probability = (13*4)(12*6)/52C5= 6/4165

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