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Math Help - William Has Two 10-Sided Spinners

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    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.
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  2. #2
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    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\%
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    Smile

    Thanks, that really helped.
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