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?

  • $\begingroup$ "drawing 5 cards" - presumably, without replacement? $\endgroup$ – Zev Chonoles Jun 27 '11 at 20:36
  • $\begingroup$ Yes, just like you would draw the first 5 cards from a regular deck of cards $\endgroup$ – Tyler Murry Jun 27 '11 at 20:41
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    $\begingroup$ The wording of your question is non-standard, and may lead to misinterpretation. (i) By "4x Red" do you mean $4$ Red? (ii) Are you asking for the probability that among the $5$ cards drawn, there will be at least one Red and at least one Green? Or at least one Red and exactly one Green? Or are you asking about the cards left in the deck? (That is what Ross Millikan's hint assumes you mean.) $\endgroup$ – André Nicolas Jun 27 '11 at 21:24

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: $$P(No Red \cup No Green) = P(No Red) + P(No Green) - P(No Red \cap No Green)$$

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  • $\begingroup$ The number of hands with exactly $1$ red and $1$ green is $4 \cdot 2 \binom{15}{3}=3640$, as you have to consider the ways of choosing the other three cards. This carries over to the other cases. $\endgroup$ – Ross Millikan Jun 27 '11 at 23:41
  • $\begingroup$ Ross, I took your comment into account. Am I on the right track? Thanks a lot for teaching me something about probability :). Also, @Tyler, thanks for letting me learn here on your question. $\endgroup$ – D. Patrick Jun 28 '11 at 0:01
  • $\begingroup$ I agree with the 5536 number. $\endgroup$ – Ross Millikan Jun 28 '11 at 0:19
  • $\begingroup$ I actually figured out how to prove it with where I was going with my original answer. Should I add it to my answer or is it generally frowned upon to edit answers here? $\endgroup$ – D. Patrick Jun 28 '11 at 2:40
  • $\begingroup$ @DPatrick, quite the opposite, editing answers is approved of, highly. $\endgroup$ – Gerry Myerson Jun 28 '11 at 3:22

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.

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  • 2
    $\begingroup$ I read it the probability that there would be at least 1 red and 1 green left among the 16. Did you mean at least 1 red and 1 green among the 5? $\endgroup$ – Ross Millikan Jun 27 '11 at 21:19

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