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Math Help - Help me with simple lotto probabillity problem?

  1. #1
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    Help me with simple lotto probabillity problem?

    Hey I was wondering if some of you maths guys could help me work something out. I have a friend who thinks he can win lotto because he can guess 18 numbers that WONíT come up and will be correct a considerable amount of times. Can you help me work out the odds for these situations?

    In Australian lotto there are 45 balls and 6 balls are drawn. What are the odds that:

    i) You pick 6 numbers and all 6 numbers are drawn

    ii) You pick 6 numbers and 5 of your 6 numbers are drawn

    iii) You pick 18 numbers that WONíT come up and all 18 numbers you choose donít come up

    iv) You pick 18 numbers that wonít come up and 17 of those 18 numbers donít come up (one does come up)

    Can you also show me how you did it?

    Thanks
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  2. #2
    MHF Contributor matheagle's Avatar
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    The number of ways that can happen is

    {45\choose 6} IF you cannot replace the balls

    Your odds are {1\over {45\choose 6}}
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  3. #3
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    Hello, deltasalt!

    In Australian lotto there are 45 balls and 6 balls are drawn.

    There are 6 Winners and 39 Others.

    There are: . \displaystyle {45\choose6} \:=\:\frac{45!}{6!\,39!} \:=\:8,\!145,\!060 possible outcomes.




    What is the probability that:

    . . i) You pick 6 numbers and all 6 numbers are drawn.

    There is one way for your 6 numbers to be drawn.

    P(\text{6 Winners}) \:=\:\dfrac{1}{8,\!145,\!060}




    ii) You pick 6 numbers and 5 of your 6 numbers are drawn.

    There are: . {6\choose5} \:=\:6 ways for 5 of your 6 numbers to be drawn.
    . . The the sixth number can be any of the 39 Others.

    Hence, there are: . 6\cdot39 \,=\,234 ways to have five Winners.

    P(\text{5 Winners}) \;=\;\dfrac{234}{8,\!145,\!060} \;=\;\dfrac{39}{1,\!357,\!510}




    iii) You pick 18 numbers that WONíT come up
    and none of your 18 numbers comes up.

    There are 18 numbers that you chose and 27 Others.

    . . \displaystyle\text{There are: }\:{27\choose6} \,=\,296,010\,\text{ to choose 6 Others.}

    P(\text{none of your 18}) \:=\:\dfrac{296,\!010}{8,\!145,\!060} \:=\:\dfrac{897}{24,\!682}




    iv) You pick 18 numbers that wonít come up,
    and exactly one of those 18 numbers comes up.

    There are 18 numbers that you chose and 27 Others.

    \displaystyle\text{There are: }\:{27\choose5} \,=\,80,\!730\text{ ways to choose 5 Others.}
    . . \displaystyle\text{and: }\:{18\choose1} \,=\,18\text{ ways to choose one of your numbers.}


    The probabiity is: . \displaystyle \frac{80,\!730\cdot18}{8,145,060} \:=\:\frac{2,\!691}{271,502}

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