Dice roll sum higher than binomial distribution

I need to find the solution for the following question:

• Person A throws 3 dice and counts the amount of threes or sixes in the result, multiplies that amount by 5 and then adds 2 to it.

• Person B throws 2 dice and counts the sum of the dice values.

Then we call X = result of person A - result of person B

Questions:

What result can we expect (so E[X])? What is the chance X>0 , X<0 and X == 0?

I started like this:

Roll of person A is a binomial distribution: Q ~ B(3,1/3) so 3 attempts with a chance of 1/3 to get success. A = 5Q + 2 <- total result of A

Chances for B to throw result (2-12) = (1/36, 2/36, ..., 6/36, ..., 2/36, 1/36) <- total result of B

Now I am stuck in this exercise and I suppose I need to make use of E[X-Y] = E[X] - E[Y]? Thank you for helping