Playing cards word problem: 5 cards are drawn from a normal deck of cards....

Shouldn't it be 12/52 * 11/51 * 10/50?
The answer is:
\(\displaystyle \Large{\dfrac{\dbinom{4}{1}\dbinom{13}{3}\dbinom{39}{2}}{\dbinom{52}{5}}}\)
That is the probability of exactly three cards of one suit in a five card hand.
 
Shouldn't it be 12/52 * 11/51 * 10/50?
The answer will only come after you seriously think about all the decisions you need to make to get three of a kind.

12/52 * 11/51 * 10/50--you are picking your 1st card that has probability of 12/52 and then you are picking your 2nd card that has probability of 11/52 and then you are picking your 3rd card that has probability of 10/52. The problem is that you at best only picked three cards.

You need to understand how when giving a deck of cards how to go about picking 5 cards that has 3 of one denomination and 2 other cards of different denomination. Random guessing will not work.

Here are the obvious steps to get 3 of a kind.

Pick a denomination (for the 3 of a kind)
Pick 3 cards of that denomination.
Now pick 2 other denominations (for the two '1 of kind')
Pick 1 card of the one of two these denominations.
Pick 1 card from the 2nd denomination.

Now that you know what the steps are you need to find out the number of ways each step can happen.
 
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The answer is:
\(\displaystyle \Large{\dfrac{\dbinom{4}{1}\dbinom{13}{3}\dbinom{39}{2}}{\dbinom{52}{5}}}\)
That is the probability of exactly three cards of one suit in a five card hand.
I was so close. But the 39C2, can it not result in them all being of same suit? I mean, could the 5 cards not all end up being hearts?
 
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The answer is:
\(\displaystyle \Large{\dfrac{\dbinom{4}{1}\dbinom{13}{3}\dbinom{39}{2}}{\dbinom{52}{5}}}\)
That is the probability of exactly three cards of one suit in a five card hand.
Prof, actually the answer is
\(\displaystyle \Large{\dfrac{\dbinom{13}{1}\dbinom{4}{3}\dbinom{12}{2}\dbinom{4}{1}^2}{\dbinom{52}{5}}}\)

The problem with \(\displaystyle \dbinom{13}{3}\) is that you are commuting the 3 cards.
 
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Could you explain those choose functions, one by one?
You really try not to listen. I spoon feed you but you don't listen.

Pick a denomination (for the 3 of a kind)--------------------------\(\displaystyle \dbinom{13}{1}\)
Pick 3 cards of that denomination.--------------------------------\(\displaystyle \dbinom{4}{3}\)
Now pick 2 other denominations (for the two '1 of kind')----------\(\displaystyle \dbinom{12}{2}\)
Pick 1 card of the one of two these denominations.----------------\(\displaystyle \dbinom{4}{1}\)
Pick 1 card from the 2nd denomination.----------------------------\(\displaystyle \dbinom{4}{1}\)
 
Prof, actually the answer is
\(\displaystyle \Large{\dfrac{\dbinom{13}{1}\dbinom{4}{3}\dbinom{12}{2}\dbinom{4}{1}^2}{\dbinom{52}{5}}}\)
You are wrong about that answer.
The original post asks for the probability of (exactly)three of the same suit.
5 cards are drawn from a normal deck of cards.
b) What are the odds that 3 of the cards were of the same suit?
See this see line for five cards. Then multiply by four.

Now see my answer.

BTW. Do you know what suit means?
 
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So pka's answer is for exactly 3 of the same suit and Jomo's answer is for at least 3 of the same suit?
 
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So pka's answer is for exactly 3 of the same suit and Jomo's answer is for at least 3 of the same suit?

I'm an idiot. My answer was for getting 3 of a kind. Like {7,7,7,Q,A} or {5,5,5,4,7). Didn't you see that in the way I made the steps. I obviously misread the problem.
I'll be sitting in the corner for awhile (forum rule!)
 
I was so close. But the 39C2, can it not result in them all being of same suit? I mean, could the 5 cards not all end up being hearts?
No. The suit that you pick the 3 cards from contains 13 cards. So after removing those 13 cards from the deck leaves 39 cards. So when you pick the two cards from the remaining 39 cards none of these 39 cards will have the suit you already chose 3 from. However, the two last cards can have the same suit!
 
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