A fair die is rolled three times, find the probability of the following events: a. All rolls show an even number of dots
b. the last two rolls show an even number of dots
c. the third roll shows an even number of dots
d. every roll shows a single dot
e. every roll shows the same number of dots
what Ive done so far:
I know that the probability will always be out of 216 because you roll the die 3 times and there are 6 possibilities. 
I think I am overthinking things. For e, would it just be 6/216 or would it be 1/216 ->( (1/6) * (1/6) * (1/6))? 
Please show me how/where you got the answers from. Thanks! 
 A: 
a. All rolls show an even number of dots

This is the probability to get an even number, multiplied by itself twice.

b. the last two rolls show an even number of dots
  c. the third roll shows an even number of dots

This concerns only the last rolls, but is quite the same thing as a.

d. every roll shows a single dot

This is the same method thing as a.

e. every roll shows the same number of dots

This is not similar. 
You can write the event e. as
$$
\bigcup_{i=1}^6
\{\omega\in\Omega : X_1(\omega) = X_2(\omega) = X_3(\omega) = i \}
$$
and sum the probabilities.
A: Reminders:  for an equiprobable sample space, $S$, you have $$Pr(A) := \frac{|A|}{|S|}$$
Your sample space in this scenario is "all possible ways of throwing three distinct six-sided dice."  (presumably the dice are fair six-sided with sides 1,2,3,4,5,6 since no other information is given)
By multiplication principle you have $6$ choices for the first die, $6$ choices for the second, and $6$ choices for the third, for a total of $6^3 = 216$ outcomes in the sample space $S$.
In each part of the question we deal with different events and try to calculate their probability.  To do this, count how many outcomes are in each event.
For (a), every roll has an even number of pips (i.e. all dice show either 2,4, or 6).  Use multiplication principle: you have three choices for the first die, three choices for the second, three choices for the third.
(b), only the final two dice show an even number of pips (the first die can be anything, the second die has to be a 2,4, or 6, and the third die has to be a 2,4, or 6).  So, use multiplication principle: six choices for the first, three choices for second, three choices for third.
(c), similarly to before, now it doesn't matter what the first or second die shows, only the third die.
(d), every die has to show a single pip (i.e. all dice show a 1).  There is only one possibility for the first die, only one for the second, and only one for the third.
(e), use multiplication principle: choose what number the first die is (six choices), the second die will be the same as the first (one choice), and the third die will be the same as the previous two (one choice).
Each of these numbers above I talked about will be the size of the event in question, so to calculate the probability, divide by $216$.  For example, part (e) you should have calculated the total number of possibilities as $6\cdot 1\cdot 1 = 6$, so the probability is $\frac{6}{216} = \frac{1}{36}$
A: Hint A: What's the probability of rolling an even, $P(even)$? You want this to happen all three times.
Hint B: We don't care about the first roll.
Hint C: We don't care about the first two rolls.
Hint D: What's the probability of rolling a 1, $P(1)$? You want this every time.
Hint E: The first roll doesn't matter. You just want to match it the second and third time.
A: Hints:
(a) Three possible even numbers out of six on each roll
(b) Consider only the last two rolls
(c) Consider only the last roll: 3/6 = 1/2
(d) How many of the 216 satisfy this very strict condition?
(e) Already done
