Solving $3$x $6$-sided dice of different colours to find the probability of events An experiment consists of $3$ fair, different coloured dice being rolled. The dice are $6$-sided and the sides show numbers $1,\dots,6$. Let $A$ be the event that none of the dice shows numbers $1$ and $2$, and let $B$ be the event that all dice show an odd number.
A) What is the probability of $A$?
B) What is the probability of $B$?
C) What is the probability of $A$ intersecting $B$?
I've solved this question by finding the total number of possible outcomes:
$|S| = 6^3 = 216$
The results were way too long for the marks given which makes me question the method I used for these solutions.
I ended up with:
A) $P(A) = \frac{64}{216} = \frac{8}{27} $ 
B) $P(B) = \frac{26}{216} = \frac{13}{108}$ 
C) $P(A \cap B) = \frac{8}{216} = \frac{1}{27}$

So to think about $A$ for the first role we have $\frac{4}{6}$ possibility as we only want $3,4,5,6$ and this is the same for the other $3$ dices, so $(\frac{4}{6})^3= \frac{8}{27}$.
Next for $B$ we have have the odd numbers to be $1,3,5$ so the possibilities are $\frac{3}{6}$ for each dice, so overall again we cube for three dices so $(\frac{3}{6})^3= \frac{1}{8}$.
Finally we have odd and not one or two, so the number on one dice can only be $3,5$ so with 3 dices: $(\frac{2}{6})^3= \frac{1}{27}$.
But your answers are correct hope this helps :)
