A very nice problem from

veritas prep that is much easier than it seems.

So, I actually drew the roulette, just to visualize it. I divided the circle in 10 sectors and wrote one number from 1 to 10 in each sector.

The question asks about the probability that in 3 spins the product of the 3 spins will be even.

What we know is that it only takes 1 even number in any multiplication for the product to result in an even number.

So, if we want at least 1 even, we can calculate the probability of not one even, or put differently all 3 spins end up in an odd number, and subtract this from 1.

5 out of 10 numbers are odd. So, the probability of odd is 5/10, which is 1/2.

We have 3 spins, so we calculate (1/2)^3 = 1/8 = 0.125 and 0.125*100 = 12.5 %.

100 - 12.5 = 87.5, so I guess

we go with A, which is the closest.