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

(C) 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. ---------- This seems incorrect. If there are said slips, 8/9 in the first draw allow us the potential to get a match in the second draw. However, only one provides a match in the second draw for each of the 8 potentials in the first draw. Therefore, the prob would be 8/9 × 1/8 = 1/9. No?

Probability questions are sometimes very tricky. Especially when we deal with some conditional outcomes. So I advise you to keep an outcome as simple as possible, when you can do so. In this question we can deal with one outcome - a pair of selected slips. And we can think of it as if the both slips were chosen at once and then we revealed the numbers. You can either deal with ordered pairs, e.i. (1, 2) and (2, 1) are different. We did so in the explanation. The probability is 14/72 = 7/36. Or you can deal with non-ordered pairs, e.i. {1, 2} and {2, 1} are the same. In this case the total number of possible outcomes is C(9, 2) = 9!/(2! × 7!) = 36 and the number of desired pairs is 7. So the probability is 7/36.

The approach, proposed by you, is also possible. However in this approach we deal with two conditional outcomes (selecting the first slip & selecting the second one). It is more complicated and there is a greater chance of making a mistake. But if you deal with it carefully you will arrive to the same result as we did in the approach above.

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This seems incorrect. If there are said slips, 8/9 in the first draw allow us the potential to get a match in the second draw. However, only one provides a match in the second draw for each of the 8 potentials in the first draw. Therefore, the prob would be 8/9 × 1/8 = 1/9. No?

The second draw is tricky here. Because if we select 1 in the first draw, the second one must be 2. However if we select 2 in the first draw, the second one can be 1 or 3. Numbers 3, 5, 8, 13, 21 also have two options for the second draw. While 34 has only one, 21. Therefore the probability is (1/9 × 1/8) + (1/9 × 2/8) + (1/9 × 2/8) + (1/9 × 2/8) + (1/9 × 2/8) + (1/9 × 2/8) + (1/9 × 2/8) + (1/9 × 1/8) = 2/9 × 1/8 + 6/9 × 2/8 = 1/36 + 6/36 = 7/36.

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