nym0112 wrote:Help! Please!

If x, y, and z are selected independently and and random from the interval [0,1], then the probability that x>=yz is...

A. 3/4
B. 2/3
C. 1/2
D. 1/3
E. 1/4

This question is from a GRE subject test in mathematics, so if you're applying to do a PhD in pure math, you might need to be able to answer it. If you're taking the GMAT, don't worry about it. The GMAT tests only discrete probability -- that is, it only tests probability in situations where there are a finite number of outcomes (flipping a coin a few times, picking marbles from a bag, picking an employee at random from a company, etc). The above question tests continuous probability -- when it mentions choosing x, y and z from the interval [0,1], that means that x, y and z could be any of the infinite number of decimals between 0 and 1 -- so you could never see it on the GMAT test.

Still, because of the answer choices, it's fairly easy to see that A must be correct; if x is not the smallest of the three numbers (i.e. if either x > y or x > z), it's absolutely certain that x > yz (since all the numbers are between 0 and 1). There's a 2/3 chance that x is not the smallest element, so the answer is at least 2/3. Since x > yz can be true even when x is the smallest element, the answer must be greater than 2/3, and A is the only answer that makes sense. Still, I can't see any way to actually prove the answer is exactly 3/4 without using calculus.