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Say you have a lotto game 10/20, which means that 10 balls are drawn from 20 which are in a cylinder.

Now, lets say I bet on one exact number to come out (for example number 6), I have the 50% chance of doing that, right?

Ok, now please tell me if I'm correct by stating that the chance of guessing three numbers (in example: i say that from 10 numbers which are drawn there will be 2, 5, 8) is this: 10/20 * 9/19 * 8/18 = 0,10 => 10%

Ok, now further more (and this one I don't know how to approach): what is the probability of me guessing exactly one number, or exactly 2 numbers, or none of them (of those three which I chose). Even further what is the probability to guess AT LEAST one number, or AT LEAST two numbers.

I will be grateful for any help, or any reading material reference.

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    @Ross: yes, he is right, so if I say that numbers 2,5,8 will be$n$that 10 numbers that are drawn then the order of them actually "coming" out is not important.2011-06-02

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If 10 out of 20 are drawn and you guess 3, then the probabilities are

  • Three correct: $\dfrac{10 \times 9 \times 8}{20 \times 19 \times 18} = \dfrac{2 }{19} \approx 0.10526 $
  • Two correct: $\dfrac{3 \times 10 \times 9 \times 10}{20 \times 19 \times 18} = \dfrac{15}{38} \approx 0.39474 $
  • One correct: $\dfrac{3 \times 10 \times 10 \times 9}{20 \times 19 \times 18} = \dfrac{15}{38} \approx 0.39474 $
  • Zero correct: $\dfrac{10 \times 9 \times 8}{20 \times 19 \times 18} = \dfrac{4 }{38} \approx 0.10526 $

As a check, those numbers add up to $1$. The factors of 3 come from the possibility of reordering correct and incorrect answers

  • At least two: $\dfrac{2 }{19} + \dfrac{15}{38} = \dfrac{1}{2} = 0.5$
  • At least one: $\dfrac{2 }{19} + \dfrac{15}{38} + \dfrac{15}{38} = \dfrac{17}{19} \approx 0.89474$
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    @ShreevatsaR: Thank you for your explanation.2011-06-03