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You have an urn with 6 numbered balls and you pull out 3 (with replacement). What is the chance of getting 3 different numbers when the order doesn't matter?

I have two solutions that both seem reasonable.

  1. Looking at chances:
    Chance to pick the first ball: 6/6
    Chance to pick the second ball: 5/6
    Chance to pick the third ball: 4/6
    i.e. 20/36

  2. Looking at possibilities (combinatorics):
    To get 3 different balls out of a urn with 6 when no number can be repeated and the order doesn't matter is the same as a lottery.

Therefore there are ${6 \choose 3}$ = 20 possibilities.
Looking at all possibilities, I have 6*6*6 = 216. i.e. 20/216

Which one is wrong and why?

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    Hint: how would the original question be different if order did matter?2011-12-17

2 Answers 2

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The $20$ possibilities are without regard to order, while the $6*6*6$ considers the order you picked the balls. This is exactly the factor $6$ between the answers. (First answer: 36; multiply by 6, equals the second answer: 216) Your $20/36$ is correct.

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    Aaah, that is my problem! :) Thank you very much!2011-12-16
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First you pick one ball.

Then you pick a second ball. With 5/6 probability it is different from the first ball.

Then you pick the third ball. With 4/6 probability it is different from the first two.

5/6 * 4/6 = 20/36.