OG Practice Test 2 - Section 6 #25

[johnhurford]

Q: Eight points are equally spaced on a circle. If 4 of the 8 points are to be chosen at random, what is the probability that a quadrilateral having the 4 points chosen as vertices will be a square?

I've read through the answer in the OG on this one however I'm not fully getting it. First time I did this problem I drew the picture and tried drawing squares and got stuck. Wondering if there's a more simple way you can explain how to plan to solve and solve this problem (e.g. how would you use annagrams? How do you know how to combine the annagrams/combo's with a probability ratio?)?

Thanks,

John

[tommywallach]

Hey John,

I'd think of it this way.

The first point chosen can be ANY point. The NEXT point chosen can be any THREE points (skipping one in either direction, or the diagonal across).

Once those points are chosen, however, the square is now defined. So there are two "correct" choices for the next position, and one for the last. We divide by the # of slots factorial as the order in which the points are chosen doesn't matter.

(8 * 3 * 2 * 1) / 4! = 4 possible squares (this could be eyeballed as well)

How many possible quadrilaterals are there? 8 * 7 * 6 * 5 / 4! = 7 * 2 * 5 = 70

4/70 = 2/35 is the answer.

Hope that helps!