Various probabilities given three dice with different colors Here is a set of probability questions. I am having trouble with the last three.
Assume that you have three six-sided dice. One is yellow (Y), one is red (R) and one is white (W). 
A)What is the probability that, in rolling only the red die, it will show the number 5?
B)What is the probability that, in rolling only the white die, it will show an even number?
C)What is the probability that, in rolling all three dice at once, the outcome will be 2(Y), 4(R), 6(W)?
D)What is the probability that, in rolling all three dice at once, the outcome will be 1, 3 and 5 in any color combination?
E)What is the probability that, in rolling all three dice at once, the outcome will be the same number on all three dice?
F)What is the probability that, in rolling all three dice at once, the outcome will be a different number on all three dice?
Here is what I have so far:
D) From part C I found that P for any # is $1/6$  thus $P(2Y,4R,6W)=(1/6)^3=1/216$. Therefore, same logic should apply here with the exception that there will be a lot more combinations w/o restriction on color. Since we have $3$ color boxes there would be total of $9$ combinations possible  and thus $P=(1/6)^9$ but its not the correct answer. On the contrary, the right answer is $P=6(1/216)=1/36$ I DO NOT UNDERSTAND why we multiply by $6$.
E) here since they ask for the same number the probability of getting say $(2,2,2)$ combo is $P= 1/6 \cdot 1/6 \cdot 1/6 = 1/216$ thus having $6$ #'s on the dice the total probability of having all the same #'s is $P = 1/216+...+1/216=6/216=1/36$
F) I dont understand as well, since we found in E that having the same #'s on all three dice is  $1/36$ then would not $P(\text{not the same #'s})= 1-1/36=35/36$? but for some reason it's not the correct answer
I would appreciate if you could explain your reasoning, I'm new to probability and pretty much forced to learn it for genetics problems.
 A: An easy way to think about (D) is that, rolling any die, you have a 3 in 6 chance of hitting a 1,3, or 5.
Then, rolling the 2nd die, you have a 2 in 6 chance of hitting one of the not-yet-attained numbers (i.e., if the 1st was a 5, of hitting 1 or 3)
Finally, the third die has a 1 in 6 chance of hitting the remaining number.
3/6 * 2/6 * 1/6 = 1/36
P.S. 1 in 216 would be the chance that a specific color (e.g., the red die) matches 1, the white matches 3, and the yellow matches 5. These numbers can be permuted in six different ways: R1W3Y5 R1Y3W5 W1R3Y5 W1Y3R5 Y1R3W5 Y1W3R5... 6 of the 216 possible unique outcomes fulfill the requirement, and 6/216 is 1/36.
Do the same thing for (E): the first die will not disqualify a roll (100% success, or 6/6). The second has a 1/6 chance of matching it, and the third also has a 1/6 chance of matching the first two.
6/6 * 1/6 * 1/6 = 1/36
For (F), it's 6/6 success for the first die, and 5/6 success for the 2nd (i.e., a 1 in 6 chance of being equal, therefore a 5 in 6 chance of being unequal). The third die has a 2 in 6 chance of being equal to any of the two already present (different-from-one-another) values, or a 4 in 6 chance of fulfilling the requirements. 6/6 * 5/6 * 4/6 = 120/216 = 20/36 = 5/9.
The reason you might be confused regarding (F) is that there are a handful of cases where neither (E) nor (F) is fulfilled. Think of the result 3,4,4, or 1,1,2... two dice the same, one different. Those are the cases not covered by E and F: 100% - 7/12 = 5/12
A: D)
In the order 1,3,5, $Pr = \dfrac{1}{6}\cdot\dfrac{1}{6}\cdot\dfrac{1}{6} = \dfrac{1}{216}$
but there could be $3! = 6$ different orders, so multiply by 6
E)
First number can be anything, each of the next two have to be different, hence $\dfrac{1}{6}\cdot\dfrac{1}{6} = \dfrac{1}{36}$
F)
6 ways to get first number, 5 ways for second one, and 4 ways for third,
hence $\dfrac{6*5*4}{216} = \dfrac{5}{9}$
Note that the complement of all same is not all different, it is all not same
