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I was given the question in class:

"What is the probability of getting a hand with 1 heart, 2 diamonds, 2 clubs?"

and

"What is the probability of getting a hand with at least 3 queens"

for the first question I assumed that the cards were removed from the deck each time so i removed them from the total each time and multiplied each probability:

13/52 * 13/51 * 12/50 * 13/49 * 12/48

but i got an abnormally small answer to this and I'm not sure if i'm on the right track..

And I'm not 100% sure what do with the second question for "atleast" questions i generally find the compliment of the event and then use that.

Could I please get some help with these two questions?

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We assume that in each problem we are picking a $5$-card hand at random.

In your first calculation, what you found is the probability of getting $1$ heart, $2$ diamonds, and $2$ clubs in that order.

The following approach will get you the right probability. There are $\binom{52}{5}$ $5$-card hands, all equally likely.

There are $\binom{13}{1}\binom{13}{2}\binom{13}{2}$ hands with $1$ heart, $2$ diamonds, and $2$ clubs. For the heart can be chosen in $\binom{13}{1}$ ways. For each choice of heart, there are $\binom{13}{2}$ ways of choosing $2$ diamonds, and for each way of choosing the heart and diamonds, there are $\binom{13}{2}$ ways to choose the clubs. Thus the required probability is $$\frac{\binom{13}{1}\binom{13}{2}\binom{13}{2}}{\binom{52}{5}}.$$


We find the number of $5$-card hands that have $3$ Queens, and the number that have $4$ Queens.

How many $3$-Queen hands are there? The Queens can be chosen in $\binom{4}{3}$ ways. For each way, we can choose the $2$ non-Queens to go with the Queens in $\binom{48}{2}$ ways. Thus there are $\binom{4}{3}\binom{48}{2}$ $3$-queen hands.

How many $4$-Queen hands are there? A similar (but simpler) calculation shows this is $\binom{4}{4}\binom{48}{1}$, or more simply $48$.

So the probability of $3$ or more Queens is $\dfrac{\binom{4}{3}\binom{48}{2}+\binom{4}{4}\binom{48}{1}}{\binom{52}{5}}$.

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  • $\begingroup$ Thank you for the reply, could you please elaborate what notation you are using for the probabilities? Maybe it's just the formatting of stackexchange: (13 1) what exactly is that and how is it calculated? $\endgroup$ – joe Mar 22 '14 at 7:22
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    $\begingroup$ You are welcome. Like most mathematicians, I use $\binom{n}{r}$ for "$n$ choose $r$", the number of ways to choose $r$ items from $n$ items. In school, it is often denoted by $C^n_r$, or $C(n,r)$, or ${}^nC_r$, or something like that. There is a general formula $\binom{n}{r}=\frac{n!}{r!(n-r)!}$. It often simplifies considerably. for example, $\binom{13}{1}=13$ and $\binom{13}{2}=\frac{(13)(12)}{2}$. It turns out that $\binom{52}{5}=\frac{(52)(51)(50)(49)(48)}{5!}$. $\endgroup$ – André Nicolas Mar 22 '14 at 7:31
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    $\begingroup$ Ohh C notation. Thanks for the help, I really appreciate it! $\endgroup$ – joe Mar 22 '14 at 7:34

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