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A box contains nine slips that are each labeled with one number: 1, 2, 17 Jul 2015, 01:28
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C
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55% (hard)

Question Stats:

66% (02:18) correct 34% (02:28) wrong based on 331 sessions

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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

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C
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A box contains nine slips that are each labeled with one number: 1, 2, 17 Jul 2015, 01:52
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Probability = No: of desired outcomes / Total No: of outcomes.

You are picking two slips out of 9 slips. So
Total No: of outcomes = 9C2 = 36

Desired outcome : sum of the numbers on the two slips is equal to one of the numbers left in the box. How many such outcomes are there?
If you look at the numbers closely, you will see that the following pair of numbers will give you the desired outcome.
(1,2) (2,3) (3,5) (5,8) (8,13) (13,21) (21,34) . There are 7 such pairs. If the two numbers which I pick is from any of these 7 pairs, then I get my desired outcome.
So No: of desired outcomes = 7

Probability = 7/36

Answer : C
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A box contains nine slips that are each labeled with one number: 1, 2, 17 Jul 2015, 01:42
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Imo c. 7/36

Only 1 and 8 have prob of 1/8 9th has prob of zero rest there are exactly two nos.one left and one right so prib 2/8. So total prob is 2/9*1/8+6/9*2/8 = 7/36
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A box contains nine slips that are each labeled with one number: 1, 2, 17 Jul 2015, 01:53
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1, 2, 3, 5, 8, 13, 21, 34 and 55.
In the series above, take any number after 2. The sum of previous 2 numbers will be equal to the number chosen.

lets take number 3 : sum of previous 2 numbers (1+2 = 3)
lets take number 55: sum of previous 2 numbers (21+34 = 55)

So in total we have 7 possibilities : (1,2) (2,3) (3,5) (5,8) (8,13) (13,21) (21,34)

probability that the sum of the numbers on the two slips is equal to one of the numbers left in the box = \(7/(9c2)\) = 7/36

Ans: C
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A box contains nine slips that are each labeled with one number: 1, 2, 17 Jul 2015, 02:48
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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

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Total number of cases = 9C2 = 36

Total favorable scenarios =

1,2
2,3
3,5
5,8
8,13
13,21
21,34

= 7

Thus the probability = 7/36, C is the correct answer.
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A box contains nine slips that are each labeled with one number: 1, 2, 19 Jul 2015, 12:44
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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

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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.
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A box contains nine slips that are each labeled with one number: 1, 2, 20 Feb 2026, 07:37
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