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July 31st, 2017, 12:18 PM  #1 
Newbie Joined: Jul 2017 From: France Posts: 1 Thanks: 0  Probability help
Hello, I am having difficulties solving this math problem. There is a deck of cards of face cards only (12 cards) where each card has 3 possible values : K, Q and J. In a hand of 5 cards, what is the P that all suits show up? What I thought about doing : For the first card, we can choose any card so p=1. The second card, there is now 11 remaining cards and I have to choose 1 suit in the remaining 3 suits hence p = 9/11 The third p = 6/10 The fourth p = 3/9 The last card can be any card so 1 again (Im probably wrong) so the total p is all of those multiplied which gives about 0.15 I think I could also do it with counting principles but i am terrible at it. Thank you so much. 
July 31st, 2017, 02:52 PM  #2 
Senior Member Joined: May 2016 From: USA Posts: 857 Thanks: 348 
If you have 5 cards and 4 suits, you must have 2 cards in one suit and 1 card in each of the other three suits, right? How many ways can you select the suit with 2 cards? Pretty obviously 4. How many ways can you select 2 cards from 3? $\dbinom{3}{2} = 3.$ In each of the other 3 suits, how many ways can you select 1 card? Obviously 3. So the number ways to pick 4 cards such that 2 cards are in one suit and 1 card in each of the others is $4 * 3 * 3 * 3 * 3 = 4 * 81 = 324.$ How many ways can you select 5 cards from 12? $\dbinom{12}{5} = \dfrac{12 * 11 * 10 * 9 * 8}{5 * 4 * 3 * 2} =$ $\dfrac{\cancel {12} * 11 * \cancel {10} * 9 * 8}{\cancel {(4 * 3)} \cancel {(5 * 2)}} = 11 * 9 * 8 = 99 * 8 = 792.$ So $p = \dfrac{324}{792} \approx 40.9\%.$ EDIT: You could do it through thinking about serial trials, but it would be much more convoluted than what you tried. I tend to avoid trying serial trials unless I have become desperate. Last edited by JeffM1; July 31st, 2017 at 02:57 PM. 
July 31st, 2017, 05:42 PM  #3  
Member Joined: Jan 2017 From: California Posts: 80 Thanks: 8  Quote:
$\frac{4\ \binom{3}{3}\ \binom{9}{2}}{\binom{12}{5}}$ Last edited by dthiaw; July 31st, 2017 at 05:45 PM.  
July 31st, 2017, 05:55 PM  #4 
Senior Member Joined: May 2016 From: USA Posts: 857 Thanks: 348  What about it? Do you have a logical explanation for that computation? I understand the 4. But after that you seem to be assuming 3 cards from 1 suit and 2 cards from the remaining suits, which adds up to 3 suits, not 4. 
July 31st, 2017, 05:56 PM  #5 
Member Joined: Jan 2017 From: California Posts: 80 Thanks: 8 
Nevermind i misread it. I understood it as the probability of all cards in a particular suit showing up

July 31st, 2017, 05:58 PM  #6 
Member Joined: Jan 2017 From: California Posts: 80 Thanks: 8  There would be 4 ways you could have all cards in one suit because of 4 suits. I misread the question though

July 31st, 2017, 06:51 PM  #7  
Member Joined: Jan 2017 From: California Posts: 80 Thanks: 8  Quote:
Last edited by dthiaw; July 31st, 2017 at 07:34 PM.  
July 31st, 2017, 07:29 PM  #8  
Senior Member Joined: May 2016 From: USA Posts: 857 Thanks: 348  Quote:
3 * 3 * 3 * 3 ways to draw 2 spades, 1 heart, 1 diamond, and 1 club, 81 in total. But there are 4 possible ways to choose the suit with 2 cards. So we get 4 * 81 = 324. That might have been a clearer explanation. These counting problems are not computationally hard, but they do require a flash of insight into structuring the problem properly.  
July 31st, 2017, 07:52 PM  #9  
Member Joined: Jan 2017 From: California Posts: 80 Thanks: 8  Quote:
 

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