William Has Two 10-Sided Spinners

• Nov 8th 2009, 01:45 PM
Bobalina
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.
• Nov 8th 2009, 01:49 PM
e^(i*pi)
Quote:

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 $\displaystyle C$ be the chance that two land the same colour and that the the two spinners are independent

$\displaystyle P'(C) = 1 - P(C)$

$\displaystyle P(C) = P(RR) + P(BB) + P(GG)$

Spoiler:
$\displaystyle P(RR) = \frac{5}{10} \times \frac{2}{10} = \frac{10}{100}$

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

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

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

$\displaystyle P'(C) = 1 - \frac{33}{100} = \frac{67}{100} = 67\%$
• Nov 9th 2009, 06:49 AM
Bobalina
Thanks, that really helped. :D