Results 1 to 11 of 11
Like Tree2Thanks
  • 1 Post By Plato
  • 1 Post By HallsofIvy

Math Help - Probability

  1. #1
    Junior Member
    Joined
    Jul 2008
    Posts
    33

    Probability

    There are 4 girls and 5 boys would like to go on school trip but there are onlyy 2 seats available.

    a) calculate the probability that one girl and one boy will be chosen
    ans P (BG)+ P(GB)=( 5/9 x 4/8 ) +( 4/9 x 5/8 )

    b) if Sarah is one of the girls who want to go on the trip, what is the probabily of not choosing Sarah?
    How to solve it?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Newbie
    Joined
    Sep 2010
    Posts
    11
    Thanks
    1

    Re: Probability

    Hi!

    Quote Originally Posted by nazz View Post
    There are 4 girls and 5 boys would like to go on school trip but there are onlyy 2 seats available.

    a) calculate the probability that one girl and one boy will be chosen
    ans P (BG)+ P(GB)=( 5/9 x 4/8 ) +( 4/9 x 5/8 )
    I don't agree with that, because you don't care about who is chosen first.
    So it's just P(one boy and one girl) = 5/9*4/8

    Quote Originally Posted by nazz View Post
    b) if Sarah is one of the girls who want to go on the trip, what is the probabily of not choosing Sarah?
    How to solve it?
    There are 9 people: sarah and 8 other students.

    P(sarah is not chosen to go on the trip) = 8/9 * 7/8

    In this case you do not care about if two boys are chosen for example. You just want to know if sarah is one of the lucky ones getting the chance to go on this trip.
    Last edited by Quant; May 1st 2012 at 07:33 AM.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,912
    Thanks
    1760
    Awards
    1

    Re: Probability

    Quote Originally Posted by nazz View Post
    There are 4 girls and 5 boys would like to go on school trip but there are onlyy 2 seats available.
    a) calculate the probability that one girl and one boy will be chosen
    ans P (BG)+ P(GB)=( 5/9 x 4/8 ) +( 4/9 x 5/8 )

    Quote Originally Posted by Quant View Post
    I don't agree with that, because you don't care about who is chosen first.
    So it's just P(one boy and one girl) = 5/9*4/8
    Actually the above answer is correct.

    It is gotten by \dfrac{\dbinom{5}{1}\dbinom{4}{1}}{\dbinom{9}{2}}=  2\cdot\frac{5}{9}\cdot\frac{4}{8}
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Newbie
    Joined
    Sep 2010
    Posts
    11
    Thanks
    1

    Re: Probability

    Hello Plato!

    Quote Originally Posted by Plato View Post
    Actually the above answer is correct.

    It is gotten by \dfrac{\dbinom{5}{1}\dbinom{4}{1}}{\dbinom{9}{2}}=  2\cdot\frac{5}{9}\cdot\frac{4}{8}
    Where does the factor 2 come from?

    I think

    \dfrac{\dbinom{5}{1}\dbinom{4}{1}}{\dbinom{9}{2}}

    is equal to
    =\frac{5}{9}\cdot\frac{4}{8}

    Am I wrong?
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Senior Member
    Joined
    Mar 2012
    From
    Sheffield England
    Posts
    440
    Thanks
    76

    Re: Probability

    9c2=36
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,912
    Thanks
    1760
    Awards
    1

    Re: Probability

    Quote Originally Posted by Quant View Post
    \dfrac{\dbinom{5}{1}\dbinom{4}{1}}{\dbinom{9}{2}}
    is equal to
    =\frac{5}{9}\cdot\frac{4}{8}
    Am I wrong?
    Yes you are.
    \dfrac{\dbinom{5}{1}\dbinom{4}{1}}{\dbinom{9}{2}}=  \dfrac{5\cdot 4}{\dfrac{9\cdot 8}{2\cdot 1}}
    Thanks from Quant
    Follow Math Help Forum on Facebook and Google+

  7. #7
    MHF Contributor

    Joined
    Apr 2005
    Posts
    16,214
    Thanks
    1789

    Re: Probability

    Quote Originally Posted by Quant View Post
    Hi!



    I don't agree with that, because you don't care about who is chosen first.
    So it's just P(one boy and one girl) = 5/9*4/8
    Actually, nazz is correct precisely because "you don't care about who is chosen first"! What you gives is the probability that a boy is chosen first and a girl second. The probability that as girl would be chosen first and then a boy would be (4/9)(5/8) and the probability of "one boy and one girl in either order" is the sum of those, as nazz gives.


    There are 9 people: sarah and 8 other students.

    P(sarah is not chosen to go on the trip) = 8/9 * 7/8

    In this case you do not care about if two boys are chosen for example. You just want to know if sarah is one of the lucky ones getting the chance to go on this trip.
    Thanks from Quant
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Newbie
    Joined
    Sep 2010
    Posts
    11
    Thanks
    1

    Re: Probability

    Quote Originally Posted by Plato View Post
    Yes you are.
    \dfrac{\dbinom{5}{1}\dbinom{4}{1}}{\dbinom{9}{2}}=  \dfrac{5\cdot 4}{\dfrac{9\cdot 8}{2\cdot 1}}
    whop, I'm terribly ashamed.
    Well, thank you!

    Quote Originally Posted by HallsofIvy View Post
    Actually, nazz is correct precisely because "you don't care about who is chosen first"! What you gives is the probability that a boy is chosen first and a girl second. The probability that as girl would be chosen first and then a boy would be (4/9)(5/8) and the probability of "one boy and one girl in either order" is the sum of those, as nazz gives.
    I see, thanks.
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Junior Member
    Joined
    Jul 2008
    Posts
    33

    Re: Probability

    What about part b) can someone shed more light on it please and thanks for all the replies
    Follow Math Help Forum on Facebook and Google+

  10. #10
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,912
    Thanks
    1760
    Awards
    1

    Re: Probability

    Quote Originally Posted by nazz View Post
    There are 4 girls and 5 boys would like to go on school trip but there are onlyy 2 seats available.
    b) if Sarah is one of the girls who want to go on the trip, what is the probabily of not choosing Sarah?
    How to solve it?
    The probability that Sarah is chosen equals \dfrac{\binom{8}{1}}{\binom{9}{2}}~.
    Follow Math Help Forum on Facebook and Google+

  11. #11
    Junior Member
    Joined
    Jul 2008
    Posts
    33

    Re: Probability

    Thanks for for all the replies you all have been great.
    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: 1
    Last Post: February 18th 2010, 02:54 AM
  4. Replies: 3
    Last Post: December 15th 2009, 07:30 AM
  5. Continuous probability - conditional probability
    Posted in the Statistics Forum
    Replies: 1
    Last Post: October 1st 2009, 02:21 AM

Search Tags


/mathhelpforum @mathhelpforum