Results 1 to 3 of 3

Math Help - sports probability

  1. #1
    Senior Member
    Joined
    Jan 2009
    Posts
    381

    sports probability

    In a sports quiz , each participant is required to answer a question , 2 of which are on badminton and 3 on soccer . THe questions are kept in separate envelopes . These envelopes are numbered 1 to 5 . The four participants are A,B C and D . They are required to pick their own questions randomly .

    For each participants , what is the probability that he will be the first to pick a question on badminton.

    so this is what i have in mind , its either he picks badminton (B) first , then 2nd guy pick B , then the rest soccers (BBSS) , or (BSBS) or (BSSS)

    there are 3 cases here , (2/5)(3/5)(3/5)(3/5)+(2/5)(3/5)(2/5)(3/5)+(2/5)(3/5)(3/5)(3/5) but the answer given is 3/10 . Erm , i cant see my mistake here .
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,914
    Thanks
    778
    Hello, thereddevils!

    The problem is not stated clearly,
    . . so my interpretation may be wrong.


    In a sports quiz, each participant is required to answer a question,
    2 of which are on badminton and 3 on soccer.
    The four participants are A, B, C and D.
    They are required to pick their own questions randomly .

    For each participant, what is the probability that he will be the first to pick a question on badminton?
    I assume the following:

    . . The participants pick one question from the five.

    . . The selected question is not replaced.

    . . The participants do their picking in random order.



    With all events being equally likely, each participant has an equal chance
    . . of being the first to choose a badminton question.

    Then why isn't this true?
    . . . P(\text{first to pick badminton}) \;=\;\frac{1}{4}

    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor
    Joined
    Dec 2009
    Posts
    3,120
    Thanks
    1
    Or, for any particular order...

    P(1st person picks B)= \frac{2}{5}=\frac{4}{10}

    P(B is picked 1st by the 2nd person)= \frac{3}{5}\frac{2}{4}=\frac{3}{10}

    P(B is first picked by the 3rd person)= \frac{3}{5}\frac{2}{4}\frac{2}{3}=\frac{2}{10}

    P(B is first picked by the 4th person)= \frac{3}{5}\frac{2}{4}\frac{1}{3}\frac{2}{2}=\frac  {1}{10}
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 10
    Last Post: January 21st 2011, 12:47 PM
  2. Replies: 3
    Last Post: May 29th 2010, 08:29 AM
  3. Replies: 2
    Last Post: January 8th 2008, 12:05 PM
  4. Fantasy Sports Algorithm
    Posted in the Advanced Statistics Forum
    Replies: 3
    Last Post: September 7th 2007, 03:11 PM
  5. Band or Sports?
    Posted in the Math Topics Forum
    Replies: 3
    Last Post: January 23rd 2007, 03:45 AM

Search Tags


/mathhelpforum @mathhelpforum