Results 1 to 4 of 4

Math Help - probability(teacher can't solve it- -)

  1. #1
    Junior Member
    Joined
    Mar 2008
    Posts
    43

    probability(teacher can't solve it- -)

    Aki is a participant on a trivia-based game show. He has an equal likelihood on any give trial of being asked a qeustion from one of six categories: Hollywood, strange places,Number Fun, Who?, Having a Ball and Write On! Aki feels that he has an 50/50 change of getting Having a Ball or Strange Places questions correct, but he thingks he has a 90% probability of getting any of the other questions right. If Aki has to get two of three questions correct, what are his odds of winning?
    The answer is 43/50, my teacher can't solve it- - what'd hell, the question sounds easy, but it relates to something i haven't learned. like binomial probability and combinations, any body help? Thank you very much.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,547
    Thanks
    539
    Hello, ives!

    I don't agree with their answer . . .


    Aki is a participant on a trivia-based game show. He has an equal likelihood on any
    given trial of being asked a question from one of six categories: A, B, C, D, E, or F.
    Aki feels that he has an 50% chance of getting an A or B question correct,
    but he thinks he has a 90% probability of getting any of the other questions right.
    If Aki has to get two of three questions correct, what are his odds ? of winning?

    The answer is 43/50 . . . . . This is not odds!

    The probability of getting an A or B question is: . \frac{2}{6} = \frac{1}{3}
    The probability of answering an A or B question correctly is: . 50\% = \frac{1}{2}
    . . Hence, the probability of answering an A or B question correctly is: . \frac{1}{3}\cdot\frac{1}{2} \:=\:\frac{1}{6}

    The probability of getting an Other question is: . \frac{4}{6} = \frac{2}{3}
    The probability of getting an Other question correct is: . 90\% = \frac{9}{10}
    . . Hence, the probability of answering an Other question correctly is: . \frac{2}{3}\cdot\frac{9}{10} = \frac{3}{5}

    So the probability of answering any question correctly is: . \frac{1}{6}+\frac{3}{5} \:=\:\frac{23}{30}
    . . The probability of getting it wrong is: . \frac{7}{30}


    There are two ways that Aki can win . . .

    (1) His first two answers are Right.
    . . P(\text{RR}) \:=\:\left(\frac{23}{30}\right)^2 \:=\:\frac{529}{900}

    (2) He answers the first one wrong, and one of the last two right.
    . . P(\text{WRW or WWR}) \:=\:2\cdot\left(\frac{23}{30}\right)^2\left(\frac  {7}{30}\right)\;=\;\frac{7,\!406}{27,\!000}


    Therefore, his probability of winning is: . \frac{529}{900} + \frac{7,\!406}{27,\!000} \;=\;{\color{blue}\boxed{\frac{5,\!819}{6,\!750}}}

    His odds of winning are approximately 25:4

    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,386
    Thanks
    1476
    Awards
    1
    Quote Originally Posted by ives View Post
    Aki is a participant on a trivia-based game show. He has an equal likelihood on any give trial of being asked a qeustion from one of six categories: Hollywood, strange places,Number Fun, Who?, Having a Ball and Write On! Aki feels that he has an 50/50 change of getting Having a Ball or Strange Places questions correct, but he thingks he has a 90% probability of getting any of the other questions right. If Aki has to get two of three questions correct, what are his odds of winning?
    The answer is 43/50,
    While I agree that the given answer cannot possibly be odds, I can show that one reading of the question gives the probability of his winning as \frac {4147}{5000} which is approx \frac {41}{50}.
    Could you have made a typo?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Junior Member
    Joined
    Mar 2008
    Posts
    43
    i don't know if the answer is right.....
    that is the answer which textbook gave out
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 1
    Last Post: February 17th 2010, 12:12 PM
  2. Replies: 4
    Last Post: December 7th 2009, 10:38 AM
  3. Is there a teacher in the house?
    Posted in the Algebra Forum
    Replies: 1
    Last Post: April 6th 2009, 04:54 PM
  4. Teacher cant even do this.......
    Posted in the Calculus Forum
    Replies: 2
    Last Post: March 31st 2008, 04:43 AM
  5. A mistake by my teacher?
    Posted in the Calculus Forum
    Replies: 6
    Last Post: December 1st 2007, 05:07 AM

Search Tags


/mathhelpforum @mathhelpforum