Probability rolling 4 dice

Question

Here is my problem, I would like to know If I did it right or not.

Rolling 4 dice

A : The face is not "two"

B : The face is not "three" or not "two"

1)What is the probability that no "two" face appears

So each time we have 5/6 chance that we dont have the "two" face

$$ P(A)^4 = (5/6)^4 = 0.4823% $$

2)What is the probability that no "two" face or no "three" face appear

Using the same logic than before, we have 4/6 chance that we dont have the "two" or the "three" face

$$ P(B)^4 = (4/6)^4 = 0.1975 $$

3) What is the probability to have 2 pairs

The probability of having distinct pairs is

$$ 15/6^4 = 0.0115 $$

I was able to find 15 by calculating it manually, but I think I find it by using combinations, can someone explain it to me.

Thank you.

Answer 1

(1)&(2) are okay.

For (3): there are $\binom 62$ ways to select two from the six faces for the pairs, $\binom 42$ ways to select among the dice to place them, and the probability that each of the four die then obtains the selected face is $(1/6)^4$, so...

$$\binom 62\binom 42\frac 1{6^4} ~=~ \dfrac{90}{6^4}~=~\dfrac 5{72}$$