This is a homework problem that I've done two different ways. The question is P(Both dice show 4 or less $\bigcap$ No dice show a 1).
I first separated them out to do P(Both dice show 4 or less) $=4/6 \cdot 4/6$ and reduced to get $4/9$.
Then I did P(No dice show a 1) = $5/6\cdot 5/6$ and got $25/36$.
Since there's a $\bigcap$ this means multiply so I took $4/9\cdot25/36$ and got $100/324 = 0.31$.
However if you simply find the number of dice combinations where both dice show 4 or less AND the number within those that neither dice shows a one you get $9/36 = 0.25$???
So, I'm trying to figure out why doing it the first way was wrong, considering there will be times when the numbers will be to large for me to count all the possible combinations.
Thanks
For each die, the probability of getting $\le 4$ is $4/6=2/3$, and the probability of getting $\ne1$ is $5/6$. It follows that the probability of getting $\le 4$ AND $\ne1$ is $2/3\times5/6$ ONLY IF those two events are independent, but they're not.
For each die, you need the probability that the outcome is in the set $\{2,3,4\}$ and that is $3/6=1/2$.
The outcomes for the two dice are independent, so you get $\dfrac12\cdot\dfrac12=\dfrac14$.