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).

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- Jul 4th 2007, 02:09 AMGAdamsDice
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 AMgalactus
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.

$\displaystyle (\frac{1}{6})(\frac{3}{6})+(\frac{1}{6})(\frac{1}{ 6})$ - Jul 4th 2007, 02:34 AMGAdams
- Jul 4th 2007, 02:50 AMCaptainBlack
The (1/6)*(3/6) is the probability of getting a 9 with a 6 on the first die and

a 3 on the second.

The (1/6)*(1/6) is the probability of getting a 0 with a 5 on the first die and

a 4 on the second.

These are mutualy exclusive outcomes so you add the probabilities.

RonL