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A 52-card deck contains 13 cards from each of the four suits: clubs , diamonds , hearts , and spades . You deal 4 cards without replacement from a well shuffled deck, so that you are equally likely to deal any 4 cards. The probability that you deal no clubs is?

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  • $\begingroup$ Welcome to math.SE: since you are new, I wanted to let you know a few things about the site. In order to get the best possible answers, it is helpful if you say in what context you encountered the problem, and what your thoughts on it are; this will prevent people from telling you things you already know, and help them give their answers at the right level. $\endgroup$ – Michael Albanese Aug 3 '14 at 16:13
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Hints:

Combinatorial way 1: There are $\binom{52}{4}$ equally likely hands of $4$. There are $\binom{39}{4}$ no club hands.

Combinatorial way 2: Imagine dealing the cards one at a time. There are $(52)(51)(50)(49)$ equally likely strings of four distinct cards. How many strings of $4$ cards have no club?

Conditional probability way: Imagine that the cards are dealt one at a time. The probability the first card is not a club is $\frac{39}{52}$.

Given that the first card is not a club, the probability the second card is not a club is $\frac{38}{51}$. So the probability that neither of the first two cards is a club is $\frac{39}{52}\cdot \frac{38}{51}$.

Given that neither of the first two card is a club, the probability the third card is not a club is $\frac{37}{50}$. Continue.

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$$\left(\frac{39}{52}\right).\left(\frac{38}{51}\right).\left(\frac{37}{50}\right).\left(\frac{36}{49}\right) = 0.303818 = 30.38\%$$

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    $\begingroup$ it would sure be nice if you added some extra comments in addition to raw numbers $\endgroup$ – mm-aops Aug 3 '14 at 17:04
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Calculate the number of combinations with no clubs:

$$\binom{39}{4}=\frac{39!}{4!\cdot35!}=\frac{36\cdot37\cdot38\cdot39}{1\cdot2\cdot3\cdot4}$$


Calculate the total number of combinations:

$$\binom{52}{4}=\frac{52!}{4!\cdot48!}=\frac{49\cdot50\cdot51\cdot52}{1\cdot2\cdot3\cdot4}$$


Divide the former by the latter:

$$\frac{\frac{36\cdot37\cdot38\cdot39}{1\cdot2\cdot3\cdot4}}{\frac{49\cdot50\cdot51\cdot52}{1\cdot2\cdot3\cdot4}}=\frac{36\cdot37\cdot38\cdot39}{49\cdot50\cdot51\cdot52}=0.303$$

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