If three distinct positive divisors of 64 are randomly selected, what is the probability that their sum will be less than 32?  Choose the correct choice from the following options: A) 1/3 B) 1/2 C) 2/7 D) 10/21 E) 15/28
The positive divisors are 1, 2, 4, 8, 16, 32, 64.   If any of the first five are chosen, the sum will be less than 32.   P(sum is less than 32) = 5/7 * 4/6 * 3/5 = 12/42 = 2/7   The answer is 2/7.
<result>C</result>
If three distinct positive divisors of 64 are randomly selected, what is the probability that their sum will be less than 32?