Bunuel
A box contains nine slips that are each labeled with one number: 1, 2, 3, 5, 8, 13, 21, 34 and 55. Two of the slips are drawn at random from the box without replacement. What is the probability that the sum of the numbers on the two slips is equal to one of the numbers left in the box?
A. 7/72
B. 1/6
C. 7/36
D. 15/36
E. 21/36
Kudos for a correct solution.
800score Official Solution:Let us consider the numbers printed on the slips i.e. {1, 2, 3, 5, 8, 13, 21, 34, 55}. There is a pattern in the numbers here, which if decoded makes the problem simpler. Starting from the third number (which is 3), each number is the sum of the previous two numbers. So: 3 = 2 + 1, 5 = 3 + 2, 8 = 5 + 3…and so on.
Now, if we want the sum on the two selected slips to be equal to one of the numbers from the set, then these two numbers have to be any two consecutive numbers from the set except the last two numbers. So, there are seven favorable pairs of numbers: {1, 2}, {2, 3}, {3, 5}, {5, 8}, {8, 13}, {13, 21} and {21, 34}. In total, there are 14 ways that these pairs can be chosen ((1, 2) and (2, 1) are two different solutions). If there are 14 possible favorable outcomes and 9 × 8 (there are 9 slips and after pulling one there are 8 left) possible outcomes, then the total probability of a favorable outcome is 14/(9 × 8) = (7 × 2)/(9 × 4 × 2) = 7 /36.
The correct answer is C.