If n dice are thrown then find the probability of the event A that the product of the numbers shown by thrown n dice is divisible by 4. I've found the following:
n(S) = 6^n
The product will be divisible if at least two of the numbers are a combination of the following:
(1 or 2 or 3 or 4 or 5 or 6,4)
or
(2,2)
or
(2,6)
I don't know how to put this information together
 A: There are precisely two disjoint ways in which the product can fail to be a multiple of 4: 
A) It could be that every number rolled is odd (i.e., is either 1 or 3 or 5).
B) It could be that all but one number rolled is odd, while the remaining value is even but not divisible by 4 (i.e., is either 2 or 6).
The probability of A is $(3/6)^n = 1/2^n$. (This is because there are $3$ ways to come out odd out of $6$ total equiprobable possibilities, on each of $n$ rolls).
The probability of B is $2/6 \times n \times (3/6)^{n - 1} = 1/3 \times n/2^{n-1} = 2/3 \times n/2^n$. (The factor of $2/6$ from the two choices out of six for the even value, the factor of $n$ for the choice of which particular roll out of $n$ total takes on the even value, and the factor of $(3/6)^{n - 1}$ for the probability of the remaining $n - 1$ rolls coming out odd, as above)
Combining these, the probability we are interested in (the probability that neither of these two disjoint events happens) is $1 - (1 + 2n/3)/2^n$.
