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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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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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