Probability of Choosing Supplementary Angles

Question

I have all of the angles from a regular triangle (60 degrees each), regular quadrilateral (90 degrees each), a regular pentagon (108 degrees each), and a regular hexagon (120 degrees each).

I have the following set of angles: 60, 60, 60, 90, 90, 90, 90, 108, 108, 108, 108, 108, 120, 120, 120, 120, 120, 120.

Now I draw two angles at random. What is the probability that they are supplementary? (answer as a reduced fraction)

I tried to do each possibility listed out, but I couldn't figure out how to go about it that way. If anyone has any recommendations or solutions, they would be much appreciated. Additionally, references to similar Stack Exchange questions would be helpful. Thank you!

Answer 1

Only way you're going to get two supplementary angles is by getting two 90s or a 60 and a 120.

(Probability two 90s) = 4/18 * 3/17 = 12/306 = 2/51

(Probability of getting a 60 and then a 120) = 3/18 * 6/17 = 18/306 = 1/17

(Probability of getting a 120 and then a 60) = 6/18 * 3/17 = 18/306 = 1/17

Total probability = 2/51 + 2/17 = 8/51

Answer 2

$$180=90+90=60+120$$

thus your probability is

$$\frac {\binom {4}{2}+\binom {3}{1}\binom {6}{1}}{\binom {18}{2} }=\frac {8}{51}$$

where $\binom {n}{p} $ is the number of sets containing $p $ elements from a set with $n $ elements . $$\binom {n}{p}=\frac {n!}{p!{}(n-p)!} $$