Please help me figure out how to do this. I'm usually okay at math but I am at a loss with this.

You draw nine cards from a standard deck of cards. What is the probability that

a. four or more will be red,

b. exactly two or three will be red,

c. two or fewer will be red, and

d. exactly five will be red?

I am assuming the 9 cards were drawn at once without replacing them.


closed as off-topic by Did, Leucippus, Sahiba Arora, Graham Kemp, TheGeekGreek Aug 7 '17 at 1:14

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  • $\begingroup$ I will do the easiest of the lot. For the last one we pick $5$ cards from the bunch of $13$ red cards, then we rest $4$ from the remaining $39$ cards. If you know basic combinatorics I think you can fill in details. $\endgroup$ – user8277998 Aug 6 '17 at 21:57
  • $\begingroup$ Answer these questions one after another: How many ways are there to draw so that all cards are red. How many ways are there to draw first five red cards then four black cards. Then how many different was are there to arrange five red cards and four black cards. How many ways are there to draw exactly fiver red cards. How many ways to draw at least five red cards. Can you try to reason those out? $\endgroup$ – fleablood Aug 6 '17 at 21:59
  • 1
    $\begingroup$ @123 I believe there are 26 total red cards, not 13. But the idea is still the same. $\endgroup$ – fleablood Aug 6 '17 at 22:01
  • $\begingroup$ @fleablood I misread red for hearts. Thanks. $\endgroup$ – user8277998 Aug 6 '17 at 22:06


The probability of $k$ out of $9$ cards being red is $$\frac{{26\choose k}{26\choose 9-k}}{{52\choose9}}$$ since there are ${52\choose9}$ different ways to select $9$ cards from a pack and ${26\choose k}$ and ${26\choose 9-k}$ number of ways to select $k$ red cards and $9-k$ black cards, respectively.


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