VJesus12 wrote:If two numbers, a and b, are to be chosen from a set of 4 consecutive integers starting with 1 and a set of three consecutive even integers starting with 4, respectively, what is the probability that b/a will not be an integer?
(A) 1/6
(B) 1/4
(C) 1/3
(D) 1/2
(E) 2/3
The OA is B. Experts, can you show me why is B the correct answer? Thanks.
We see that a is chosen from the set {1, 2, 3, 4} and b is chosen from the set {4, 6, 8}. Since there are 4 numbers in the first set and 3 numbers in the second set, the total number of ordered pairs (a, b) is 4 x 3 = 12. Of these ordered pairs (a, b), we have (1, 4), (1, 6), (1, 8), (2, 4), (2, 6), (2, 8), (3, 6), (4, 4), and (4, 8) that produce an integer when b is divided by a. integer. In other words, the probability that b/a is an integer is 9/12 = 3/4, and thus the probability that b/a is not an integer is 1 - 3/4 = 1/4.
Answer: B
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