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xEnOnn

  • 4 years ago

Suppose there is a set of numbers: \(\{1,2,3,4,5\}\). I pick from the set of numbers 3 times. Each pick is independent. So there is replacement. Then, what is the probability of me getting a 1, a 3 and a 5? And, what is the probability of getting a 2, two 4s?

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  1. myko
    • 4 years ago
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    1/125 i would say

  2. xEnOnn
    • 4 years ago
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    This is for the first part or the second part of the question?

  3. myko
    • 4 years ago
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    same

  4. myko
    • 4 years ago
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    for bouth

  5. xEnOnn
    • 4 years ago
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    why would it be the same for the second one? There is a repeated number in there ie, two 4s.

  6. myko
    • 4 years ago
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    but there is a replacement. So it doesn't mater....

  7. xEnOnn
    • 4 years ago
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    Actually, I was thinking why wouldn't it be \(\frac{6}{125}\) for the first part, and \(\frac{3}{125}\) for the second part?

  8. Zarkon
    • 4 years ago
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    you should let people know if the order that you pick then numbers matters.

  9. xEnOnn
    • 4 years ago
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    @Zarkon I think the orders do matter in this case since it is a combination because I just need to get a 1, a 3, and a 5. Similarly for the other combination of 2,4,4.

  10. Zarkon
    • 4 years ago
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    if the order matter than myko's answer is correct. If the order does not matter then you are correct. whether the order matters or not depends on what the problem is asking for. to me, it sounds like the order does not matter...though the problem could have been written better

  11. xEnOnn
    • 4 years ago
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    oh yea you are right that the order does not matter. I said it wrongly in my previous comment. Thanks for the help! :)

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