Results 1 to 7 of 7

Math Help - help on a probability problem

  1. #1
    ilc
    ilc is offline
    Newbie
    Joined
    Feb 2010
    Posts
    7

    help on a probability problem

    Two cards are drawn from a deck of 52 plating cards. find the probability of each of the following events occurring:

    a) both cards are clubs
    b) both cards are red
    c) both cards are queens
    d) both cards are red queens
    e) both cards are queens or both cards are red
    -----------

    a) 13 clubs out of 52 = 13/52
    there will be 12 clubs left after picking the first one so: 12/52
    so, (13/52)x(12/52) = 156/2704 chances to have both cards as clubs? is this correct?

    b) (26/52)x(23/52) = 650/2704

    C) (4/52)x(3/52) = 12/2704

    d) (26/52)x(23/52) = 2/2704


    how would i do e?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    -1
    e^(i*pi)'s Avatar
    Joined
    Feb 2009
    From
    West Midlands, England
    Posts
    3,053
    Thanks
    1
    Quote Originally Posted by ilc View Post
    Two cards are drawn from a deck of 52 plating cards. find the probability of each of the following events occurring:

    a) both cards are clubs
    b) both cards are red
    c) both cards are queens
    d) both cards are red queens
    e) both cards are queens or both cards are red
    -----------

    a) 13 clubs out of 52 = 13/52
    there will be 12 clubs left after picking the first one so: 12/52
    so, (13/52)x(12/52) = 156/2704 chances to have both cards as clubs? is this correct?

    b) (26/52)x(23/52) = 650/2704

    C) (4/52)x(3/52) = 12/2704

    d) (26/52)x(23/52) = 2/2704


    how would i do e?
    Where did you get 23/52 from in part B? Surely removing 1 red card would mean that 25 remain?

    Is the first card replaced? I have assumed it is not

    All your second terms should have a denominator of 51.

    For example in B:

    \frac{26}{52} \times \frac{25}{51} = \frac{25}{102}

    And in C:

    \frac{4}{52} \times \frac{3}{51} = \frac{1}{221}


    Then part E would be

    \frac{25}{102} + \frac{1}{221}
    Follow Math Help Forum on Facebook and Google+

  3. #3
    ilc
    ilc is offline
    Newbie
    Joined
    Feb 2010
    Posts
    7
    yes, that should be 25...that was a typo. thanks!
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    Joined
    Dec 2009
    Posts
    3,120
    Thanks
    1
    Quote Originally Posted by ilc View Post
    Two cards are drawn from a deck of 52 plating cards. find the probability of each of the following events occurring:

    a) both cards are clubs
    b) both cards are red
    c) both cards are queens
    d) both cards are red queens
    e) both cards are queens or both cards are red
    -----------

    a) 13 clubs out of 52 = 13/52
    there will be 12 clubs left after picking the first one so: 12/52
    so, (13/52)x(12/52) = 156/2704 chances to have both cards as clubs? is this correct?

    b) (26/52)x(23/52) = 650/2704

    C) (4/52)x(3/52) = 12/2704

    d) (26/52)x(23/52) = 2/2704


    how would i do e?
    Hi ilc,

    in making these calculations, you are replacing the first card.
    This is incorrect as you are choosing 2 from 52,
    hence after picking the first card, you have 51 left.

    Therefore redo parts a), b) and c).

    In part d), you must count the number of red queens.

    For part e),

    You can calculate the probabilities of getting 2 red cards and add to the probability of getting 2 queens....

    but you must subtract the probability of getting 2 red queens as this is a conditional probability question. They will have already been accounted for.
    There is overlap since some queens are red.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    ilc
    ilc is offline
    Newbie
    Joined
    Feb 2010
    Posts
    7
    how would i get the probability of the two red queens?

    2/53 x 1/51 = 2/2652 correct?
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor
    Joined
    Dec 2009
    Posts
    3,120
    Thanks
    1
    You only need to know how a pack is organised.

    There are 26 red and 26 black.

    There are 4 queens, 2 are red and 2 are black.

    Therefore 2 of the 26 red cards are red queens.

    Therefore there are 24 red cards that are not queens.

    There are another 2 queens which are black.

    If both cards are queens or both are red, then we must include the probability of also getting 2 black queens.
    Also, the queens can be a black and red.
    However if we include the 2 red queens, we must be aware that these are already counted,
    since the 2 red queens have been already included in 2 red cards total.

    Hence we can add the probabilities of 2 red cards to the probability of getting 2 queens,
    but we must subtract the probability of getting 2 red queens.
    Last edited by Archie Meade; February 7th 2010 at 11:28 AM. Reason: additional sentences
    Follow Math Help Forum on Facebook and Google+

  7. #7
    MHF Contributor
    Joined
    Dec 2009
    Posts
    3,120
    Thanks
    1
    Quote Originally Posted by ilc View Post
    how would i get the probability of the two red queens?

    2/53 x 1/51 = 2/2652 correct?
    \frac{2}{52}\ \frac{1}{51}

    Or choose both red queens at once.... \frac{\binom{2}{2}}{\binom{52}{2}}
    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. Probability problem
    Posted in the Statistics Forum
    Replies: 1
    Last Post: January 14th 2010, 08:45 PM
  3. Replies: 0
    Last Post: October 8th 2009, 09:45 AM
  4. probability problem
    Posted in the Statistics Forum
    Replies: 3
    Last Post: April 6th 2009, 09:09 AM
  5. Probability Problem
    Posted in the Statistics Forum
    Replies: 13
    Last Post: February 5th 2009, 04:29 PM

Search Tags


/mathhelpforum @mathhelpforum