Here are the positive divisors of 64: {1, 2, 4, 8, 16, 32, 64}AAPL wrote:Veritas Prep
If two distinct positive divisors of 64 are randomly selected, what is the probability that their sum will be less than 32?
A. 1/3
B. 1/2
C. 5/7
D. 10/21
E. 15/28
OA D
We can see that, as long as we DON'T select the 32 or 64, then the sum of the two numbers is guaranteed to be less than 32
So, P(sum is less than 32) = P(neither is a 64 nor the 32 is selected)
= P(no 64 or 32 is selected on the 1st try AND no 64 or 32 is selected on the 2nd try)
= P(no 64 or 32 is selected on the 1st try) x P(no 64 or 32 is selected on the 2nd try)
= 5/7 x 4/6
= 20/42
= 10/21
Answer: D
Cheers,
Brent
