# Probability of Choosing Supplementary Angles [closed]

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!

## closed as off-topic by Namaste, Claude Leibovici, user91500, hardmath, John BJun 5 '17 at 13:31

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Namaste, Claude Leibovici, user91500, hardmath, John B
If this question can be reworded to fit the rules in the help center, please edit the question.

• Your Question is not explicit about whether the "two angles" draw "at random" are sampled with replacement or without replacement. It is possible that you overlooked this detail in trying to "do each possibility listed out", so it might fall into place once you realize that the first choice and the second choice each have only four possible outcomes. – hardmath Jun 5 '17 at 12:41
• I couldn't figure that out either. I assume it's without replacement, and am doing the problem as such. Thank you for your clarification. @hardmath – Johnny Jun 5 '17 at 14:53

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

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

$$\frac {\binom {4}{2}+\binom {3}{1}\binom {6}{1}}{\binom {18}{2} }=\frac {8}{51}$$
where $\binom {n}{p}$ is the number of sets containing $p$ elements from a set with $n$ elements . $$\binom {n}{p}=\frac {n!}{p!{}(n-p)!}$$