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- Nov 4th 2008, 11:13 AMndcruzDice Probability
- Nov 4th 2008, 11:26 AMjoksy
The propability is not my strength but, I remember that we have 1/36 to have 2 and 1/36 to have 12 as overall outcome.

so for not 2 could be $\displaystyle 1 - \frac{1}{36}$

for 4 $\displaystyle \frac{3}{36}$

and for 6 $\displaystyle \frac{5}{36}$

=> think to 2 you have 1/6 for the first dice and 1/6 for the second so 1/36

Hope that it is useful (Sleepy) - Nov 4th 2008, 11:32 AMSoroban
Hello, ndcruz!

It's just matter of, isn't it?__counting__

There are 36 possible outcomes.

There are: .$\displaystyle 1 + 3 + 5 \:=\:9$ ways to get a sum of 2, 4 or 6.

So there are: .$\displaystyle 36 - 9 \:=\:27$ ways for some other sum.

Therefore: .$\displaystyle P(\text{sum not 2, 4, or 6}) \;=\;\frac{27}{36} \;=\;\frac{3}{4} $