# Math Help - William Has Two 10-Sided Spinners

1. ## 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.

2. Originally Posted by Bobalina
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:
$
P(RR) = \frac{5}{10} \times \frac{2}{10} = \frac{10}{100}$

$
P(BB) = \frac{3}{10} \times \frac{7}{10} = \frac{21}{100}$

$
P(GG) = \frac{2}{10} \times \frac{1}{10} = \frac{2}{100}$

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

$P'(C) = 1 - \frac{33}{100} = \frac{67}{100} = 67\%$

3. Thanks, that really helped.