1. ## 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).

2. 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.

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

3. 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

Thanks, the first part is clear 1/6 * 3/6 but I don't understand why you are adding the 1/6 * 1/6
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