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This is a continuation on a question I asked a few years back:

Say you have a 60 card deck containing 12 red cards and 48 black cards. After drawing 7 cards, what is the probability you will have 2 or more red cards?

I've tried to apply a modification to the answer's logic of the aforementioned question, but with no luck.

Community
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Tyler Murry
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1 Answers1

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Well, you need to find the probability of getting one red card, and getting zero red cards.

The chance of getting 2 or more red cards is 1-probability of getting one or zero red cards.

Getting zero red cards would be $\frac{48}{60}\frac{47}{59}...\frac{42}{54}$ the chance that each of your 7 draws is black.

The chance of getting one red card can not be found using your previous question. It seems as if it is the case that a red card is x, and there are 12 of them, but that is the chance of at least one. You need exactly one. Can you think of a formula to find the chance of exactly one?

If you can find that, the odds of 2 or more red cards is equal to (1 - (odds of zero red) - odds of one red))

Asimov
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  • Edit: added comment on relation to first question – Asimov Oct 25 '14 at 23:04
  • Based on the following: http://substepr.com/w/index.php?title=Find_the_probability_of_a_player_being_given_exactly_one_ace_in_a_Bridge_deal, I believe I have found the solution. This way is also applicable to finding the answer to my previous question much more easily. If you would like to post the full answer, I will mark it as such. – Tyler Murry Oct 25 '14 at 23:24
  • If I've helped you solve the question, then there is no need to edit. My work is done when you understand, and elaborating on something that you know is redundant. – Asimov Oct 25 '14 at 23:47
  • Marked as answer. I figured it would give other users more clarity if you elaborated in your answer. – Tyler Murry Oct 26 '14 at 00:41