Results 1 to 3 of 3

Math Help - Lottery Probability

  1. #1
    Newbie
    Joined
    Mar 2008
    Posts
    2

    Lottery Probability

    I'm having a tough time with this problem and was hoping someone coule help.

    It involves the Mega Millions lotter.

    5 numbers are selected from the intergers 1-56 (without replacement) and the special 6th ball is selected from a new set of number 1-46. A winning ticket must have correctly picked the 5 numbers from the first set and the 1 number from the 6th set.

    I know the probability of winning with 1 ticket is
    1/175711536

    The question is, if exactly $200,000,000 is wagered (at $1 per ticket) what is the probability of 2 or fewer winners? And what is the expected number of winners?

    Any help would be greatly appreciated.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    GAMMA Mathematics
    colby2152's Avatar
    Joined
    Nov 2007
    From
    Alexandria, VA
    Posts
    1,172
    Awards
    1
    Quote Originally Posted by reyup View Post
    I'm having a tough time with this problem and was hoping someone coule help.

    It involves the Mega Millions lotter.

    5 numbers are selected from the intergers 1-56 (without replacement) and the special 6th ball is selected from a new set of number 1-46. A winning ticket must have correctly picked the 5 numbers from the first set and the 1 number from the 6th set.

    I know the probability of winning with 1 ticket is
    1/175711536

    The question is, if exactly $200,000,000 is wagered (at $1 per ticket) what is the probability of 2 or fewer winners? And what is the expected number of winners?

    Any help would be greatly appreciated.
    BTW, I believe your probability of winning one ticket is wrong. It should be equal to the fraction of one over 56*55*54*53*52*46 = 21,085,384,320. Let that be equal to p.

    Probability of two or fewer can be broken down as P(2), P(1) and P(0).

    P(2)= \frac{200,000,000!}{2!*(200,000,000-2)!}*p^2 * (1-p)^{200,000,000-2}

    P(1) = \frac{200,000,000!}{1!*(200,000,000-1)!}*p \Rightarrow 200,000,000p*(1-p)^{200,000,000-1}

    P(0) = \frac{200,000,000!}{0!*(200,000,000-0)!}*p \Rightarrow p(1-p)^{200,000,000-0}

    The answer to your question is the sum of those three probabilities.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    4
    Quote Originally Posted by reyup View Post
    I'm having a tough time with this problem and was hoping someone coule help.

    It involves the Mega Millions lotter.

    5 numbers are selected from the intergers 1-56 (without replacement) and the special 6th ball is selected from a new set of number 1-46. A winning ticket must have correctly picked the 5 numbers from the first set and the 1 number from the 6th set.

    I know the probability of winning with 1 ticket is
    1/175711536

    The question is, if exactly $200,000,000 is wagered (at $1 per ticket) what is the probability of 2 or fewer winners? And what is the expected number of winners?

    Any help would be greatly appreciated.
    You are expected to use the Poisson approximation to the Binomial distribution for this problem.


    RonL
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. lottery probability
    Posted in the Advanced Statistics Forum
    Replies: 0
    Last Post: March 2nd 2011, 07:44 AM
  2. Probability, lottery...
    Posted in the Advanced Statistics Forum
    Replies: 2
    Last Post: April 27th 2010, 06:52 AM
  3. Lottery Probability
    Posted in the Statistics Forum
    Replies: 1
    Last Post: March 5th 2010, 08:07 AM
  4. Lottery probability
    Posted in the Advanced Statistics Forum
    Replies: 1
    Last Post: April 29th 2009, 10:20 PM
  5. Lottery Probability
    Posted in the Advanced Statistics Forum
    Replies: 3
    Last Post: November 18th 2007, 04:31 AM

Search Tags


/mathhelpforum @mathhelpforum