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 July 31st, 2017, 11:18 AM #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, 01:52 PM #2 Senior Member   Joined: May 2016 From: USA Posts: 1,047 Thanks: 430 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. Thanks from dthiaw and mimi3108 Last edited by JeffM1; July 31st, 2017 at 01:57 PM.
July 31st, 2017, 04:42 PM   #3
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Quote:
 Originally Posted by JeffM1 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.
$\frac{4\ \binom{3}{3}\ \binom{9}{2}}{\binom{12}{5}}$

Last edited by dthiaw; July 31st, 2017 at 04:45 PM.

July 31st, 2017, 04:55 PM   #4
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 Originally Posted by dthiaw How about $\frac{4\ \binom{3}{3}\ \binom{9}{2}}{\binom{12}{5}}$
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, 04:56 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 Thanks from JeffM1
July 31st, 2017, 04:58 PM   #6
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 Originally Posted by JeffM1 And the logic behind that computation is what? 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.
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, 05:51 PM   #7
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 Originally Posted by JeffM1 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.
Something is bugging me JeffM1. School me. Shouldnt the number of ways to choose one card from the other suits be 3 factorial. So the numerator should look like $4\times \binom{3}{2}\times 3\times \binom{3}{1}$

Last edited by dthiaw; July 31st, 2017 at 06:34 PM.

July 31st, 2017, 06:29 PM   #8
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 Originally Posted by dthiaw Something is bugging me JeffM1. School me. Shouldnt the number of ways to choose one card from the other suits be 2 since one card would have been already counted. So 2 multiplied by 3. So that the numerator becomes $4\times 3\times 3\times 2$
Consider a suit, spades for example, from which we select 2 cards. There are 3 ways to do that selection from spades. Then there are 3 ways to select 1 card from the 3 remaining hearts, 3 ways to select 1 card from the remaining 3 diamonds, and 3 ways to select 1 card from the 3 remaining clubs. So that means that there are
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, 06:52 PM   #9
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 Originally Posted by JeffM1 Consider a suit, spades for example, from which we select 2 cards. There are 3 ways to do that selection from spades. Then there are 3 ways to select 1 card from the 3 remaining hearts, 3 ways to select 1 card from the remaining 3 diamonds, and 3 ways to select 1 card from the 3 remaining clubs. So that means that there are 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.
Finally got it. awesome. Thanks a bunch JeffM1