Results 1 to 6 of 6

Math Help - A rather hard probability question.

  1. #1
    Newbie
    Joined
    Sep 2007
    Posts
    3

    A rather hard probability question.

    Hi there.

    The question is:

    A bag contains eight counters. Five of the counters are green, three are red. A counter is taken at random from the bag and its colour is noted. The counter is not replaced. A second counter is then taken from the bag. Calculate the probability that:

    (i) the first counter is green and the second is red
    (ii) the counters are of the same colour
    I've done (i), it was pretty basic. But (ii) has me stumped. Anyone got any ideas?

    Thanks in advance,

    - Chris Down
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Bar0n janvdl's Avatar
    Joined
    Apr 2007
    From
    Meh
    Posts
    1,630
    Thanks
    6
    Quote Originally Posted by Chris Down View Post
    Hi there.

    The question is:

    I've done (i), it was pretty basic. But (ii) has me stumped. Anyone got any ideas?

    Thanks in advance,

    - Chris Down


    I'm not really good with probability, so excuse me if these are not completely correct.

    Five of the counters are green, three are red.
    For 1)

    We have 8 counters in total, the chances of drawing a green counter would be 5 out of 8, which is 62,5 %

    When the 2nd counter is drawn, there are only 7 counters left.

    Chances of drawing a red counter would then be 3 out of 7, which is a chance of 42,9%

    The chances that the first one is green AND the second one is red is:

    \frac{5}{8} \times \frac{3}{7}

    = \frac{15}{56}

    = 26,8 %


    -------------------

    For 2)

    The chances of drawing two green counters are:

    \frac{5}{8} \times \frac{4}{7}

    = \frac{5}{14}

    = 35,7 %



    The chances of drawing two red counters are:

    \frac{3}{8} \times \frac{2}{7}

    = \frac{3}{28}

    = 10,7 %


    So the chances of drawing either a green pair, or a red pair is:

    \frac{5}{14}  + \frac{3}{28}

    = \frac{13}{28}

    = 46,4 %
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Sep 2007
    Posts
    3
    Quote Originally Posted by janvdl View Post
    The chances of drawing two green counters are:

    \frac{5}{8} \times \frac{4}{7}

    = \frac{5}{14}

    = 35,7 %



    The chances of drawing two red counters are:

    \frac{3}{8} \times \frac{2}{7}

    = \frac{3}{28}

    = 10,7 %
    Sorry to bother you again, but I am unsure as to how you have got the:

    \frac{4}{7}

    and

    \frac{2}{7}...

    How were these numbers acquired?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    is up to his old tricks again! Jhevon's Avatar
    Joined
    Feb 2007
    From
    New York, USA
    Posts
    11,663
    Thanks
    3
    Quote Originally Posted by janvdl View Post
    I'm not really good with probability, so excuse me if these are not completely correct.



    For 1)

    We have 8 counters in total, the chances of drawing a green counter would be 5 out of 8, which is 62,5 %

    When the 2nd counter is drawn, there are only 7 counters left.

    Chances of drawing a red counter would then be 3 out of 7, which is a chance of 42,9%

    The chances that the first one is green AND the second one is red is:

    \frac{5}{8} \times \frac{3}{7}

    = \frac{15}{56}

    = 26,8 %


    -------------------

    For 2)

    The chances of drawing two green counters are:

    \frac{5}{8} \times \frac{4}{7}

    = \frac{5}{14}

    = 35,7 %



    The chances of drawing two red counters are:

    \frac{3}{8} \times \frac{2}{7}

    = \frac{3}{28}

    = 10,7 %


    So the chances of drawing either a green pair, or a red pair is:

    \frac{5}{14}  + \frac{3}{28}

    = \frac{13}{28}

    = 46,4 %
    your reasoning is correct, i didn't check the computations. so if those check out, you're good
    Follow Math Help Forum on Facebook and Google+

  5. #5
    is up to his old tricks again! Jhevon's Avatar
    Joined
    Feb 2007
    From
    New York, USA
    Posts
    11,663
    Thanks
    3
    Quote Originally Posted by Chris Down View Post
    Sorry to bother you again, but I am unsure as to how you have got the:

    \frac{4}{7}
    we had 8 counters to begin with. the chances of choosing a green one is \frac 58. having chosen a green counter, there are now 7 counters left, 4 of which are green, since you took out one of the green ones. now the chances of choosing another green counter is \frac 47



    and

    \frac{2}{7}...

    How were these numbers acquired?
    a similar reasoning holds here
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Newbie
    Joined
    Sep 2007
    Posts
    3
    Quote Originally Posted by Jhevon View Post
    we had 8 counters to begin with. the chances of choosing a green one is \frac 58. having chosen a green counter, there are now 7 counters left, 4 of which are green, since you took out one of the green ones. now the chances of choosing another green counter is \frac 47
    Thanks a lot man. I owe you one.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Hard Probability Question
    Posted in the Advanced Statistics Forum
    Replies: 3
    Last Post: August 3rd 2010, 08:15 AM
  2. Probability Question- could be simple/could be hard??
    Posted in the Advanced Statistics Forum
    Replies: 1
    Last Post: May 7th 2010, 04:56 AM
  3. I'm stuck with this hard probability question
    Posted in the Statistics Forum
    Replies: 1
    Last Post: December 11th 2009, 12:35 PM
  4. hard probability question
    Posted in the Statistics Forum
    Replies: 5
    Last Post: June 14th 2009, 04:51 AM
  5. Hard probability
    Posted in the Statistics Forum
    Replies: 0
    Last Post: October 15th 2007, 12:32 AM

Search Tags


/mathhelpforum @mathhelpforum