Keep making these kinds of videos, it is very helpful.

An easier way to do it is to calculate the diagonal of the square. X^2 = (6 root 3)^2 + (6 root 3)^2. Solving for X leads to ~14.7. The shortest side of the square is the side of the square with 6 root 3, and the longest side is the diagonal. Since 12 is closer to 6 root 3, the probability is less than 50%. Hence, B!

Could this be done with the angles? ▫️ 90 degree angle at the corner of the square ▫️ the 30 degrees from upper left triangle contain lesser & equal lengths ▫️ the 30 degrees from lower right triangle contain lesser & equal lengths ▫️ Total of 60 degrees (2/3 of 90) contain the lesser & equal lengths ▫️ The 30 degrees (1/3 of 90) left over in-between contains the greater lengths ▫️ 1/3 is less than 50 percent.

Probability of having a distance more than 12 should be equal to 1 - the probability of having any distance ranging from 1 to 12. So getting the probability of having 12 as a distance should be about 0.56 and taking that away from 1 will give us the probability of having a distance greater than 12 which is quantity A, and equals 0.44 or 44% and is definitely less than Quantity B. So B is greater.

Thank you. Where are you getting these questions? Is this from the actual exam?

@testprepstudios

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Thanks for tuning in! This is an original question I created myself