Two fair dice rolled 5 Times Two fair dice are rolled 5 times. Let the random variable x represent the number of times that the sum 12 occurs. The table below occurs describes the probability distribution. Find the value of the missing probability.
x | P(x)
0   0.868615786121484
1   0.124087969445926
2   0.00709074111119579
3   0.000202592603177022
4   ????????????????????
5   1.65381716879202e-08

Been trying to do this for a good half hour. My attempts were:
1) Assuming it adds up 1, so I added up all the P(x) and took that away 
from 1.
2) Figuring out the probability you sum a 12 is 1/36 (6,6) with two dice. 
Hence the probability you get it 4 times is (1/36)^4.

What am I missing?
 A: Per @BGM's Comment, $X \sim \mathsf{Binom}(5, 1/36).$ You can apply
the formula for the PDF of a binomial distribution to obtain the
probabilities in the given table. 
In R statistical software, the computation looks like this:
x = 0:5;  pdf = dbinom(x, 5, 1/36)
cbind(x, pdf)
#  x          pdf
# 0 8.686158e-01
# 1 1.240880e-01
# 2 7.090741e-03
# 3 2.025926e-04
# 4 2.894180e-06
# 5 1.653817e-08


Now you can use the binomial PDF formula to obtain $P(X=4)$ and perhaps
a few of the other probabilities in this distribution.
Thank you for showing what you have tried. That always makes it easier to
be helpful. Here are some comments on your proposed methods:
Of course, all of these probabilities must add to $1$ as you say, so
you could also use that method to verify the missing entry.
But your computation $(1/36)^4 =  5.953742e-07$ is incorrect because
it does not account for the one roll out of five that doesn't have
a total of 12--or the fact that the 'non-12' could happen on any of the
five rolls. Compare that computation with the binomial formula, and maybe
you'll see what I mean.
