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

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
$$P(a_2+b_2+c_2>0, a_3+b_3+c_3>0) = 1-P(a_2+b_2+c_2=0\ or \ a_3+b_3+c_3=0)$$
$$ = 1-P(a_2+b_2+c_2=0)-P(a_3+b_3+c_3=0)+P(a_2+b_2+c_2=0, a_3+b_3+c_3=0)$$
$$ (by\ independence)= 1-P(a_2=0)^3-P(a_3=0)^3+P(a_2=0,a_3=0)^3$$
$$ = 1-P(1,3,\ or \ 5)^3-P(1,2,4, \ or \ 5)^3+P(1 \ or \ 5)^3$$
$$ = 1 - (\frac{3}{6})^3 - (\frac{4}{6})^3+(\frac{2}{6})^3 = \frac{133}{216}.$$
