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Probability Calculation - Lottery

  1. Feb 22, 2010 #1
    1. The problem statement, all variables and given/known data

    To win a lottery, must pick 5 different numbers from the 45 available.

    The order in which the numbers are chosen does not matter.

    With only one ticket, what is the probability of winning (i.e. matching all 5 numbers drawn with all 5 chosen) ?

    2. Relevant equations

    Stated within the solution

    3. The attempt at a solution

    n = number of elements in the field (in this case, 45)
    p = number of choices (5)

    [tex]P(win) = \left(\frac{n!}{(p!(n - p)!)}\right)[/tex]

    Therefore:

    [tex]=\left(\frac{45!}{(5!(45- 5)!)}\right)[/tex]


    [tex]=\left(\frac{45!}{(5!)(40!)}\right)[/tex]


    [tex]=\left(\frac{45!}{(120)(40!)}\right)[/tex]


    [tex]=\left(\frac{45!}{(5!(40)!)}\right)[/tex]


    [tex]= \left(\frac{45*44*43*42*41}{120}\right)[/tex]


    [tex]=1221759[/tex]

    Therefore:

    [tex]P(win)= \left(\frac{1}{1221759}\right) \approx 8.18\times10^{-7}[/tex]

    .. Is this correct method / answer?
     
    Last edited: Feb 22, 2010
  2. jcsd
  3. Feb 22, 2010 #2

    kuruman

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    [tex]
    \frac{45!}{(5!)(40!)}=\frac{45*44*43*42*41*40!}{5!*40!}=\frac{45*44*43*42*41}{5!}
    [/tex]

    Is there a question here?
     
  4. Feb 22, 2010 #3
    .. Just edited my initial post to make it a bit clearer and to correct some errors. Wanted to know if this was the correct method for calculating the probability?, and hence the correct answer?
     
  5. Feb 22, 2010 #4

    kuruman

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    It is correct.
     
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