+0

# Probability

+1
49
2

In a regular icosagon (20-sided polygon), all the diagonals are drawn. If a diagonal is selected at random, what is the probability of selecting a diagonal that is either the shortest possible length, the second shortest possible length, or the third shortest possible length? Express your answer as a common fraction.

Jul 8, 2022

#1
0

Probability = (Number of successful outcomes)/(Number of Total Outcomes)

The number of total outcomes is how many diagonals there are in total. 20(17)/2 is the number of diagonals which is equal to 170.

The shortest diagonal comes from connecting one vertex to another that is only two "sides" away. The amount of these diagonals is  20.

Therefore, the answer is 20/170 = 2/17.

Edit: Below whoops, didn't see that they also needed second shortest and third shortest... In that case, we'll have to multiply 20 by 3 for the other cases as well giving us

6/17.

Jul 8, 2022
edited by Voldemort  Jul 8, 2022
edited by Voldemort  Jul 8, 2022
edited by Voldemort  Jul 8, 2022
#2
+1

There are $$20 \times (20 -3) \div 2 = 170$$ diagonals.

Note that there are 20 diagonals that are the shortest(1 vertice apart), 20 diagonals that are the second shortest(2 vertices apart), and 20 diagonals that are the third shortest(3 vertices apart).

Thus, the probability is $${{3 \times 20} \over 170 } = {\color{brown}\boxed{6 \over 17}}$$

Jul 8, 2022