Probability of Drawing a Particular set of Cards from a Deck I think this is a pretty simple question.
If I've got a deck composed of the following cards:
4x Red, 2x Green, 4x Blue, 8x Orange, 3x Purple
What is the probability, after drawing 5 cards, that there will be at least 1 Red card and 1 Green card?
 A: I'm assuming you mean, if you draw 5 cards of the 21 you listed, what's the probability that your hand will contain at a minimum of 1 red card and a minimum of 1 green card.
If that's the case, I believe this is the solution:
If you have 21 choose 5 cards (because order doesn't matter), there are 20,349 possible hands.
$$\binom{21}{5}=20,349$$
So, look at the odds of getting the most basic winning hand: 1 red and 1 green.
There are $\binom{2}{1} = 2$ ways to get 1 green and $\binom{4}{1} = 4$ ways to get 1 red (green 1 + red 1, green 1 + red 2, etc.).  Then, there are $\binom{15}{3} = 455$ ways to select from the remaining non-green non-red cards.
Thus, there are $\binom{2}{1} \cdot \binom{4}{1} \cdot \binom{15}{3} = 3640$ ways to get a winning hand with 1 green, 1 red, and 3 non-green non-reds.  If we add that to the other possible hands, we'll know the total number of winning combinations.
The possible winning hands and their counts are:


*

*3640 hands with 1 green, 1 red, and 3 other  

*420 hands with 2 green, 1 red, and 2 other   

*1260 hands with 1 green, 2 red, and 2 other  

*90 hands with 2 green, 2 red, and 1 other    

*120 hands with 1 green, 3 red, and 1 other   

*4 hands with 2 green, 3 red, and 0 other 

*2 hands with 1 green, 4 red, and 0 other


You have 5,536 possible winning combinations.  The probability of getting one of those hands is 5,536 in 20,349 or .272.

One other approach ...
Consider the following:
$$!Red = \binom{17}{5} = 6,188$$
$$!Green = \binom{19}{5} = 11,628$$
Each of those hands would result in a winning hand (i.e., either no greens or no reds).  The problem is that they both include those hands which have neither greens nor reds so we have to the duplicates out.
$$!(Red | Green) = \binom{15}{5} = 3,003$$
If we add the hands without red to the hands without green and take out the duplicates we get $6,188 + 11,628 - 3,003 = 5,536$, the same as before.
Basically, I believe the following is true:
$$P(No Red \cup No Green) = P(No Red) + P(No Green) - P(No Red \cap No Green)$$
A: Hint: To not have 1 red and 1 green left, you have to draw all 4 reds or draw both greens.  Since you can't draw 4 reds and 2 greens (6 cards) these are disjoint and can be added.
