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The packet of 10 cards includes 4 aces and cards: 5,6,7,8,9 and 10.

What is the probability that:

  1. Four randomly chosen cards are consecutive numbers.
  2. In the four of randomly chosen cards isn't any ace.
  3. In the four of randomly chosen cards are two aces and two consecutive numbers.
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Is this homework? If so, you should tag it as such. There are many people who try to use this site to just do their homework for them. By being open and honest, I think you'll receive a better response. If you don't know how to retag, let me know and I'll do it for you. –  Quinn Culver Apr 22 '12 at 13:28
    
I don't know, how does it work here. It is a task about probability. It isn't homework. I found this in some test. After half day of counting and reading articles about combinatorics, I'm not able to solve this. Especially I can't figure out how to count all posibilities. –  Annmn Apr 22 '12 at 13:37

1 Answer 1

up vote 2 down vote accepted

Hints: There are $\binom{10}{4}$ ways to choose a "hand" of $4$ cards from the $10$. All hands are equally likely. So to answer the three questions, all we need to do is to count the hands that qualify, and divide each count by $\binom{10}{4}$. The numbers are small, we could count by making careful lists.

$1.$ How many hands have $4$ consecutive cards? Surely we can list them all, without any theory. There is $\{5,6,7,8\}$, and $\{6,7,8,9\}$, and $\{7,8,9,10\}$.

$2.$ How many aceless hands are there? We need to choose $4$ cards from the $6$ non-aces.

$3.$ We can choose the $2$ Aces in $\binom{4}{2}$ ways. For each of these ways, the two consecutives can be chosen in $5$ ways.

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Thank you. You are right. –  Annmn Apr 22 '12 at 14:00

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