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I have 10 cards, 5 of them are red and other 5 are green. What's the probability of drawing 3 red cards out of 5 draws?

What I have so far is ((5*4*3*5*4)/5!)/(10*9*8*7*6)

Which is ~ 0.033% I'm pretty sure it's not correct. So could you please point out the error I've made and correct it?

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Do you assume the cards of the same color to be identical (then better call them balls) or do they have other values? –  Dennis Gulko Jan 24 '12 at 18:46
    
Perhaps if, instead of dividing by $5!=120$, you multiplied by ${5 \choose 3}=10$ you might get closer. –  Henry Jan 24 '12 at 18:49
    
@Dennis: it should not make any difference –  Henry Jan 24 '12 at 18:49
    
@Henry Doing so I got ~0.397% In reality it's pretty easy to draw 3 red cards, so this seems wrong, it probably should be around 30% I think. –  Cobold Jan 24 '12 at 18:56
    
@Cobold:No - it is a probability of about 0.397 which is 39.7% –  Henry Jan 24 '12 at 19:00
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2 Answers

up vote 2 down vote accepted

Upstairs it seems you are counting unordered outcomes and downstairs it seems you are counting ordered outcomes. This is not right, you need to be consistent...

I'll solve the problem both ways.

With ordered outcomes and thinking of all cards as distinct:

The total number of ordered outcomes is $10\cdot9\cdot 8\cdot 7\cdot 6$.

The number of ordered outcomes in which exactly 3 cards are red is:

${5\choose3} \cdot( 5\cdot4\cdot 3 )\cdot 5\cdot4$.

The $5\choose 3$ is the number of ways to choose which of the three draws are red (it seems you had an error here). Then the number of ways to choose the cards for the "red slots" is $5\cdot4\cdot3$ and the number of ways to choose the other two cards is $5\cdot4$.

So the probability is $${{5\choose3} \cdot( 5\cdot4\cdot 3 )\cdot 5\cdot4\over10\cdot9\cdot 8\cdot 7\cdot 6 } ={ 10\cdot 25\cdot16\cdot3\over 10\cdot 9\cdot 8\cdot 7\cdot 6 }={25\over63}.$$


With unordered outcomes:

There are $10\choose 5$ total outcomes and ${5\choose3}\cdot{5\choose2}$ desired outcomes. So the probability is $${{5\choose3}\cdot{5\choose2} \over {10\choose 5}} ={25\over63}.$$

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3 red out of 5: $\dfrac{5 \times 4 \times 3}{3 \times 2 \times 1} \times \dfrac{5 \times 4 \times 3 \times 5 \times 4}{10 \times 9 \times 8 \times 7 \times 6 } \approx 0.3968$ as the reds and greens can come in any order.

4 red out of 5: $\dfrac{5 \times 4 \times 3 \times 2}{4 \times 3 \times 2 \times 1} \times \dfrac{5 \times 4 \times 3 \times 2 \times 5}{10 \times 9 \times 8 \times 7 \times 6 }$

5 red out of 5: $\dfrac{5 \times 4 \times 3 \times 2 \times 1}{5 \times 4 \times 3 \times 2 \times 1} \times \dfrac{5 \times 4 \times 3 \times 2 \times 1}{10 \times 9 \times 8 \times 7 \times 6 }$

Add those up and you get $\frac{1}{2}$. Also add 0, 1 or 2 reds out of 5 and you get $1$.

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