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Math Help - Probability with Combinatorics

  1. #1
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    Probability with Combinatorics

    if the lottery had 49 numbers you could choose, and you have to pick 6, what are the chances that:

    1. All your numbers are chosen
    2. 5 of your numbers are chosen
    3. 4 of your numbers are chosen

    for #1, i did 1/(49C6), is that the correct way to do it? my reasoning behind it is that there's 49C6 combinations one could choose, and only 1 set of those are what you have, and what could get picked.

    i'm really confused for questions 2 and 3, so any help would be appreciated thx
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  2. #2
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    Hello, mneox!

    If the lottery had 49 numbers you could choose, and you have to pick 6,
    what are the chances that:

    1. All your numbers are chosen
    2. 5 of your numbers are chosen
    3. 4 of your numbers are chosen

    For #1, i did 1/(49C6), is that the correct way to do it?
    My reasoning behind it is that there's 49C6 combinations one could choose,
    and only 1 set of those are what you have, and what could get picked.
    Right!
    There are: . _{49}C_6 possible outcomes.


    2. Exactly five of your numbers are chosen.

    Among the 49 possible numbers, 6 are "winners", the other 43 are "losers".

    We want 5 winners (from the available 6 winners)
    . . and 1 loser (from the available 43 losers).

    There are: . (_6C_5)(_{43}C_1) ways.

    Therefore: . P(\text{exactly 5 winners}) \;=\;\frac{(_6C_5)(_{43}C_1)}{_{49}C_6}


    Now you know how to solve #3.

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  3. #3
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    Thank you, your answers were very informative and I got the answers now!
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