Results 1 to 4 of 4

Math Help - Probabilities and Odds Problems

  1. #1
    Newbie
    Joined
    Apr 2008
    Posts
    19

    Probabilities and Odds Problems

    Hey,

    Can someone help me on this plz and explain your solution.

    A confident and boastful coach claims that on the next game the odds of her team winning are 3:1, the odds against losing are 5:1, and the odds against tieing are 7:1. Can these odds be right? Show why or why not these claims make sense.

    Thanx

    Ballin
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Joined
    Apr 2008
    Posts
    1,092
    Quote Originally Posted by ballin_sensation View Post
    Hey,

    Can someone help me on this plz and explain your solution.

    A confident and boastful coach claims that on the next game the odds of her team winning are 3:1, the odds against losing are 5:1, and the odds against tieing are 7:1. Can these odds be right? Show why or why not these claims make sense.

    Thanx

    Ballin
    I find it easiest to work these problems by rewriting the odds as probabilities.
    The probability of winning is 3/4. Similarly, the probability of not losing is 5/6, and the probability of not tying is 7/8.
    But the probability of tying should be the probability of not losing minus the probability of winning, which we find is:
    \frac{5}{6} - \frac{3}{4} = \frac{20}{24} - \frac{18}{24} = \frac{2}{24} = \frac{1}{12},
    whence the probability of not tying is 11/12. So you would conclude that these odds do not make sense.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Apr 2008
    Posts
    19
    Thanx bro i get it. One more thing these probabilities are supposed to add up to 1 right?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    Joined
    Apr 2008
    Posts
    1,092
    Quote Originally Posted by ballin_sensation View Post
    Thanx bro i get it. One more thing these probabilities are supposed to add up to 1 right?
    Let P(W) represent the probability of winning;
    let P(T) represent the probability of tying;
    let P(L) represent the probability of losing.

    Then P(W) + P(T) + P(L) = 1.

    Note, however, that you are not explicitly given P(T) or P(L), though both P(T) and P(L) are easily computed from your given information.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 3
    Last Post: September 22nd 2011, 10:46 AM
  2. Approximate joint probabilities using marginal probabilities
    Posted in the Advanced Statistics Forum
    Replies: 0
    Last Post: June 23rd 2011, 05:04 AM
  3. bookies odds vs real odds
    Posted in the Statistics Forum
    Replies: 10
    Last Post: February 18th 2011, 05:49 PM
  4. Odds and probabilities
    Posted in the Statistics Forum
    Replies: 1
    Last Post: June 22nd 2009, 01:52 PM
  5. Statistics Probabilities Problems
    Posted in the Statistics Forum
    Replies: 1
    Last Post: March 29th 2009, 02:02 PM

Search Tags


/mathhelpforum @mathhelpforum