Propability of drawing at least 5 cards of the same suit? From a deck of 52 cards (the standard deck) I get 7 cards and I want to draw at least 5 cards of the same suit. What is the probability of this event?
I tried to solve this by saying that this probability equals to:
$\dfrac{{13\choose5}.{39\choose2}+{13\choose6}.{39\choose1}+{13\choose7}}{{52\choose7}} \approx 0,76\%$
That's because from the 13 cards of the same suit, we choose exactly 5 (and 2 from the rest 39 cards remain) or exactly 6 (and 1 remains) or exactly 7 of the same suit.
After this, I had to find the probability of getting 3 pairs of any number and 1 card from the remaining ones. There may be 2 pairs of the same number, meaning four of a kind plus one pair plus one card.
I think there are: 
$$26*25*24*46=717600$$ ways for that because at first, from the 26 pairs we choose 1 then from the 25 remaining we choose again 1 then from the 24 remaining we choose 1 and from the 46 cards remaining we choose the last card.
So the posibility of this is $\approx 0,5\%$
Both of the probabilities seem to be too small and I am not sure if the way I calculated them is the right one. Can someone tell me where am I mistaken?
Thanks in advance!
 A: 
Find the probability that if seven cards are drawn from a standard deck that at least five of them are of the same suit.

You correctly calculated the number of ways of choosing at least five cards from a particular suit.  However, we also have to choose one of the four suits.  Hence, the probability is 
$$\binom{4}{1}\left[\frac{\binom{13}{5}\binom{39}{2} + \binom{13}{6}\binom{39}{1} + \binom{13}{7}\binom{39}{0}}{\binom{52}{7}}\right]$$

Find the probability that if seven cards are drawn from a standard that three pairs and one other cards are drawn.

There are thirteen ranks from which the three pairs must be drawn.  For each of the $\binom{13}{3}$ selections of the ranks of the pairs, we can choose two of the four cards of that rank in $\binom{4}{2}$ ways.  This leaves us with $10$ ranks from which we can choose the remaining card and four ways of choosing a card from that rank.  Hence, the probability is 
$$\frac{\binom{13}{3}\binom{4}{2}^3\binom{10}{1}\binom{4}{1}}{\binom{52}{7}}$$
Note:  I am using the standard interpretation of three pairs, which is that the three pairs are of different ranks.  If you allow four of a kind, you could count the same hand six ways since there are $\binom{4}{2}$ ways of selecting two of the pair to constitute a pair.
A: the correct answer to your 2nd question is:
[13*C(4,2) + 12* C(4,2) + 11*C(4,2]*40/3!/C(52,7)

