Probability: Cards 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?
 A: 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. 
A: $$\left(\frac{39}{52}\right).\left(\frac{38}{51}\right).\left(\frac{37}{50}\right).\left(\frac{36}{49}\right) = 0.303818 = 30.38\%$$
A: 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:
$$\frac{\frac{36\cdot37\cdot38\cdot39}{1\cdot2\cdot3\cdot4}}{\frac{49\cdot50\cdot51\cdot52}{1\cdot2\cdot3\cdot4}}=\frac{36\cdot37\cdot38\cdot39}{49\cdot50\cdot51\cdot52}=0.303$$
