If you roll one die, you can always make a sum of 7 if you get the perfect roll on the second die. There's a 1/6 chance you roll the perfect number, so that's the probability of getting a sum of 7. The probability we do that twice in a row is (1/6)(1/6) = 1/36.

You could get down to A or B instantly by noticing that the answer needs to be the square of some very simple fraction of integers.

AAPL wrote:Manhattan Prep

What is the probability that the sum of two dice will yield a 7, and then when both are thrown again, their sum will again yield a 7? assume that each die has 6 sides with faces numbered 1 to 6.

A. \(1/144\)
B. \(1/36\)
C. \(1/12\)
D. \(1/6\)
E. \(1/3\)

OA B

Since there are a total of 36 outcomes, and 6 of them yield a sum of 7 (notice that these 6 outcomes are: {1, 6}, {2, 5}, {3, 4}, {4, 3}, {5, 2} and {6, 1}), the probability that the two numbers on the dice yield a sum of 7 is 6/36 = 1/6. The probability that their sum will again yield a 7 is 1/6. Therefore, the probability that the two dice will yield a sum of 7 twice is 1/6 x 1/6 = 1/36.

Answer: B