probability of rolling 3 dice numbers which product is a multiple of 6 As the question states, I need to find the probability of rolling 3 dice numbers which product is a multiple of 6.
what I have tried:
$\omega={(6,k,k), (3,2,k), (3,4,k))}$
$P(A) = \frac{3\times(1^1\times6^2) + 6\times(1^1\times1^1\times6^1) + 6\times(1^1\times1^1\times6^1)}{6^3}=\frac{180}{216}=\frac{5}{6}$
I know writing $1^1$ seems silly, but just trying to show what I'm doing
Thanks for helping and sorry if it is a bad question
 A: Let the dice be thrown in sequence or otherwise be distinguishable some other way. There are $6^3 = 216$ total different equally likely outcomes.
Approach by trying to count how many of these outcomes were "bad" and should be removed because they either are not multiples of $2$, are not multiples of $3$, and keeping in mind that they might have been both.

 Of these, $3^3=27$ are "bad" because their product of the results is not even (and thus not a multiple of six), seen by noting a product is odd iff all terms in the product is odd, there being three odd numbers.

$~$

Similarly, $4^3=64$ are "bad" because their product of results is not a multiple of three (and thus not a multiple of six), seen by noting a product is not a multiple of three iff all terms in the product are not multiples of three, there being four possible non-multiple of three numbers to choose from here.

$~$

 Finally, some of these "bad" outcomes we counted twice as they were simultaneously not even and not multiples of three.  These would be the outcomes consisting only of $1$'s and $5$'s, there being $2^3=8$ of which.  We keep this in mind to correct our count as we proceed with inclusion-exclusion.

The probability then is:

$$\frac{6^3-3^3-4^3+2^3}{6^3} = \frac{216 - 27 - 64 + 8}{216} = \frac{133}{216}$$

A: Let $a_2,b_2,c_2$ ($a_3,b_3,c_3$) denote the exponents of $2$ ($3$) in the prime factorization of the number shown on the three dice. Then the probability you want is
$$P(a_2+b_2+c_2>0, a_3+b_3+c_3>0) = 1-P(a_2+b_2+c_2=0\ or \ a_3+b_3+c_3=0)$$
$$ = 1-P(a_2+b_2+c_2=0)-P(a_3+b_3+c_3=0)+P(a_2+b_2+c_2=0, a_3+b_3+c_3=0)$$
$$ (by\ independence)= 1-P(a_2=0)^3-P(a_3=0)^3+P(a_2=0,a_3=0)^3$$
$$ = 1-P(1,3,\ or \ 5)^3-P(1,2,4, \ or \ 5)^3+P(1 \ or \ 5)^3$$
$$ = 1 - (\frac{3}{6})^3 - (\frac{4}{6})^3+(\frac{2}{6})^3 = \frac{133}{216}.$$
