# Binomial Coefficient Probability Question

completely stuck in this probability question. I know to use Hypergeometric probability but im not sure about what numbers i should be using. Any help would be great. A regular deck of 52 playing cards has 13 ranks in 4 suits. Jack, Queen and King of each suit are face cards. Suppose you are randomly dealt seven cards. What is the probability of getting

(b) Three face cards in the same suit and any four cards in another suit (but all four in the same suit)?

(c) Three face cards not all in the same suit, and any four non-face cards?

• Are you able to do (b) ? Aug 11, 2015 at 13:41
• No im not quite sure what numbers to use for (b) or (c), I know the method as I used it for (a) but I'm really stuck on (b) and (c) Aug 11, 2015 at 13:45

## 2 Answers

Let's start with a hint for (b) and then pause awhile

(Choose suit for face cards)(choose face cards)(choose another suit)(choose 4 cards from it)/ ${52\choose 7}$

$${{4\choose 1}{3\choose3}{3\choose1}{13\choose4}\over{52\choose 7}}$$

Of course, this could have been simplified, e.g. we could have simply counted 4 ways to select all 3 face cards from a suit, but I deliberately gave it step by step.

With the hint given on (c) by drhab, you should be able to do (c) now

• I managed to answer (a) "Four face cards and three non-face cards?" by doing (12C4)(40C3) / (52C7), but for (b) im really unsure of what numbers to use. I'm not quite sure i understand the hint tho? Thanks for taking the time to help Aug 11, 2015 at 14:05
• Ah okay I now see what you meant by your original hint. This was quite close to the solution i was working on, thanks so much for the help. Aug 11, 2015 at 14:33
• You're welcome :) Aug 11, 2015 at 14:35

Hint on c)

There are $\binom{12}3$ possibilities to choose $3$ face cards. There are $4$ possibilities to choose $3$ face cards all in the same suit.

So there are $\binom{12}3-4$ possibilities to choose $3$ face cards not all in the same suit.

• ah okay I think i understand, iv'e never used minus in this context before tho. Would (12C3)x(4C1) be correct for face card part? And then produce an equation of (12C3)x(4C1)x(40C4) / (52C7) for the final answer? Aug 11, 2015 at 14:38
• The final answer on c) is $\frac{((12C3)-4)\times 40C4}{52C7}$ Aug 11, 2015 at 14:46
• Okay thanks a lot for all the help. One last thing, I have another question: "Probability of 7 non face cards all in the same suit", would the final answer to this be (4C1) x (10C7) / (52C7) ? Thanks again. Aug 11, 2015 at 14:58
• Yes, that is correct. You are welcome. Aug 11, 2015 at 14:59