Results 1 to 3 of 3

Thread: William Has Two 10-Sided Spinners

  1. November 8th 2009, 01:45 PM #1
    Bobalina
    Bobalina is offline
    Newbie
    Joined
    Sep 2009
    From
    England
    Posts
    6

    William Has Two 10-Sided Spinners

    William has two 10-sided spinners.
    The spinners are equally likely to land on each of their sides.

    Spinner A has 5 red sides, 3 blue sides and 2 green sides.
    Spinner B has 2 red sides, 7 blue sides and 1 green side.

    William spins spinner A once.
    William spins spinner B once.

    Work out the probability that spinner A and spinner B do not land on the same colour.
    Follow Math Help Forum on Facebook and Google+
  2. November 8th 2009, 01:49 PM #2
    e^(i*pi)
    e^(i*pi) is offline
    -1
    e^(i*pi)'s Avatar
    Joined
    Feb 2009
    From
    West Midlands, England
    Posts
    3,053
    Thanks
    1
    Quote Originally Posted by Bobalina View Post
    William has two 10-sided spinners.
    The spinners are equally likely to land on each of their sides.

    Spinner A has 5 red sides, 3 blue sides and 2 green sides.
    Spinner B has 2 red sides, 7 blue sides and 1 green side.

    William spins spinner A once.
    William spins spinner B once.

    Work out the probability that spinner A and spinner B do not land on the same colour.
    Let C be the chance that two land the same colour and that the the two spinners are independent

    P'(C) = 1 - P(C)

    P(C) = P(RR) + P(BB) + P(GG)

    Spoiler:
    <br /> P(RR) = \frac{5}{10} \times \frac{2}{10} = \frac{10}{100}

    <br /> P(BB) = \frac{3}{10} \times \frac{7}{10} = \frac{21}{100}

    <br /> P(GG) = \frac{2}{10} \times \frac{1}{10} = \frac{2}{100}

    P(C) = \frac{10}{100} + \frac{21}{100} + \frac{2}{100} = \frac{33}{100}


    P'(C) = 1 - \frac{33}{100} = \frac{67}{100} = 67\%
    Follow Math Help Forum on Facebook and Google+
  3. November 9th 2009, 06:49 AM #3
    Bobalina
    Bobalina is offline
    Newbie
    Joined
    Sep 2009
    From
    England
    Posts
    6

    Smile

    Thanks, that really helped.
    Follow Math Help Forum on Facebook and Google+
« Some help required please | What am I doing wrong with this normal probability distribution? »

Similar Math Help Forum Discussions

  1. 10 Sided Die
    Posted in the Statistics Forum
    Replies: 7
    Last Post: October 16th 2010, 06:07 PM
  2. William Feller Textbook:Introduction to Probability
    Posted in the Advanced Statistics Forum
    Replies: 8
    Last Post: November 23rd 2007, 09:41 AM
  3. 8 sided die
    Posted in the Advanced Statistics Forum
    Replies: 1
    Last Post: April 5th 2007, 01:48 PM
  4. Six-sided Die
    Posted in the Statistics Forum
    Replies: 5
    Last Post: January 23rd 2007, 05:47 PM

Search Tags


/mathhelpforum @mathhelpforum