Bunuel wrote:
If Griffin rolls a 12-sided die (numbered 1- 12) 7 times, what is the probability that the sum of the results of the 7 rolls will be even?
A. 11/32
B. 1/2
C. 5/8
D. 11/16
E. 13/16
Using symmetry, we can see that probability of getting an even sum is 1/2 and of getting an odd sum is 1/2, just like it works with 6 sided dice rolled 3 times.
We know how the roll of 3 regular dice (1-6) work.
Lowest sum possible = 1 + 1 + 1 = 3 (odd) in 1 way only
Highest sum possible = 6 + 6 + 6 = 18 (even) in 1 way only
Second lowest sum possible (2 + 1 + 1) = 4 (even) in 3 ways
Second highest sum possible (5 + 6 + 6) = 17 (odd) in 3 ways
....
So probability of getting an even sum is 1/2 and of getting an odd sum is 1/2.
This case of 12 sided die rolled 7 times is no different.
Lowest sum possible = 1 + 1 + 1+...+ 1 = 7 (odd) in 1 way only
Highest sum possible = 12 + 12 + 12 +...+12 = 84 (even) in 1 way only
Second lowest sum possible (2 + 1 + 1+...+1) = 8 (even) in 7 ways (7 rolls are distinct)
Second highest sum possible (11 + 12 + 12+ ... + 12) = 83 (odd) in 7 ways (7 rolls are distinct)
...
Same logic here.
Answer (B)