Chance of getting six in three dice

Question

I am having a hard time wrapping my head around this and am sure that my answers are wrong.

There are three dice.

A. Chance of getting exactly one six on the three dice. $$(1/6) * 3 = 1/3$$

B. Chance of getting exactly two sixes. $$(1/6 * 1/6) * 1.5 = 1/24$$

C. Chance of getting exactly $~3~$ sixes. $$1/6 * 1/6 * 1/6 = 1/216$$

D. Chance of any combination of A, B and C $$1/3 + 1/24 + 1/216= 72/216 + 9/216 + 1/216 = 82/216$$

Answer 1

A. There is a total of 6^3=216 combinations if you roll 3 dice. There are 5^2x3=75 combinations that you will get one 6. Thus there is a 75/216=25/72 chance of getting only one 6 when rolling 3 dice.

B. There are 5x3 combinations that you will get 2 6s. Thus there is a 15/216=5/72 chance of getting a 2 6s when rolling 3 dice.

C. There is 1 combination where you will get 3 6s. Thus there is a 1/216 chance you will get 3 6s when rolling 3 dice. (Good job you got this correct)

D. There is a 75+15+1/216=91/216 chance of any of them happening.

Answer 2

Hint for A: what is the chance that no sixes appear? If you know that chance then you automatically know the chance that sixes do appear.

Answer 3

The chance of getting at least one 6 with three dice is 91/216 because if you subtract 125/216 (the probability of rolling three dice without getting a 6) from 216/216 (the probability of any combination of numbers), you get 91/216.