Hi LUANDATO,

We're told that a roulette wheel is divided into 10 equal sectors, with each sector bearing a distinct integer from 1 to 10, inclusive. Each time the wheel is spun, a ball randomly detemines the winning sector by setting in that sector. The wheel is spun three times and we're asked for the approximate probability that the PRODUCT of the three winning sectors' integers will be EVEN. This question is based on some Number Properties and Probability math.

To start, the product of any integer and an EVEN number is EVEN. Thus, if any of the 3 numbers is EVEN, then the product will be EVEN. Rather than calculate all of the different outcomes that involve at least one even number, we can instead calculate what we DON'T WANT to have happen (the product is ODD); we can then subtract that probability from the number 1 to determine the probability of what we DO want (the product is EVEN).

With the integers 1-10 inclusive, we have 10 total values (5 odds and 5 evens). The probability of the three numbers all being ODD is....

(Odd)(Odd)(Odd) = (5/10)(5/10)(5/10) = (1/2)(1/2)(1/2) = 1/8

Thus, the probability of AT LEAST one of the three numbers being EVEN is 1 - 1/8 = 7/8

1/8 = .125 = 12.5%
7/8 = .875 = 87.5%

The approximate value of our calculation is 88%

Final Answer: A

GMAT assassins aren't born, they're made,
Rich