# Dice

• Jul 4th 2007, 02:09 AM
Dice
I worked out that (a) the chances of the same score is 1/6 for the ordinary die and 4/6 for the other die.

Independent events, so multiplied together gives 1/9.

I am having problems with (b).
• Jul 4th 2007, 02:19 AM
galactus
It appears there are only two ways to get a sum of 9. The ordinary die has a 6, then the other must have a 3. The ordinary die has a 5, then the other must have a 4.

$(\frac{1}{6})(\frac{3}{6})+(\frac{1}{6})(\frac{1}{ 6})$
• Jul 4th 2007, 02:34 AM
Quote:

Originally Posted by galactus
It appears there are only two ways to get a sum of 9. The ordinary die has a 6, then the other must have a 3. The ordinary die has a 5, then the other must have a 4.

$(\frac{1}{6})(\frac{3}{6})+(\frac{1}{6})(\frac{1}{ 6})$

Thanks, the first part is clear 1/6 * 3/6 but I don't understand why you are adding the 1/6 * 1/6
• Jul 4th 2007, 02:50 AM
CaptainBlack
Quote: