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! 

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
