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The only contents of a container

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Manager
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The only contents of a container [#permalink] New post 08 Nov 2019, 08:15
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Question Stats:

61% (01:56) correct 38% (01:17) wrong based on 18 sessions
The only contents of a container are 14 disks that are each numbered with a different integer that is one of the integers 1 through 14 inclusive. If 3 disks are randomly removed from the container, one after the other, and without replacement, what is the probability that the median of the numbers on the 3 selected disks will be 8?

A) \(\frac{3}{26}\)
B) \(\frac{3}{14}\)
C) \(\frac{2}{7}\)
D) \(\frac{1}{3}\)
E) \(\frac{165}{364}\)
[Reveal] Spoiler: OA
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Re: The only contents of a container [#permalink] New post 08 Nov 2019, 08:19
Again for the Million time to fall in the probability trap. This question I can't figure out why the possible outcomes were found through combinations and my way didn't work! what is my pitfall?

when I solved it like this : (7/14) * (1/13) * (6/12) = 42/2184

the numerator forms the required outcomes and the den. the possible outcome . pls help. My trials to overcome probability question aren't succeeding
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Re: The only contents of a container [#permalink] New post 08 Nov 2019, 11:13
Expert's post
There are GRE exams that will not contain a probability question at all or maybe only one.

So I would not stress too much on this area
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Re: The only contents of a container [#permalink] New post 23 Nov 2019, 14:27
I also feel the same as you guys.

Even though we get stuck, we can still find the right answer.
If 3 disks are to be picked up randomly, the probability would be 3/14.
By adding additional, the probability to chose 3 disks with 8 as a median has to be less than 3/14.
Fortunately, only one answer fulfills that condition.
So, the answer is A.
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Re: The only contents of a container [#permalink] New post 24 Nov 2019, 07:36
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Expert's post
Asmakan wrote:
The only contents of a container are 14 disks that are each numbered with a different integer that is one of the integers 1 through 14 inclusive. If 3 disks are randomly removed from the container, one after the other, and without replacement, what is the probability that the median of the numbers on the 3 selected disks will be 8?

A) \(\frac{3}{26}\)
B) \(\frac{3}{14}\)
C) \(\frac{2}{7}\)
D) \(\frac{1}{3}\)
E) \(\frac{165}{364}\)


Let's solve this question using counting methods.
In order for the median (of the three disks) to be 8, one of the disks must be 8, one must be less than 8, and the other must be greater than 8.
I how many ways can we have such a configuration?

We can select the 8 in 1 way
We can select a number less than 8 in 7 ways (1,2,3,4,5,6, or 7)
We can select a number greater than 8 in 6 ways (9,10,11,12,13 or 14)
By the Fundamental Counting Principle (FCP), the total number of ways to get a median of 8 = (1)(7)(6) = 42

Now let's determine the TOTAL number of possible outcomes when we choose 3 discs.
Since the order in which we select the three discs does not matter, we can use COMBINATIONS
We can select 3 of the 14 discs in 14C3 ways
14C3 = (14)(13)(12)/(3)(2)(1) = 364

So, P(median is 8) = 42/364 = 3/26

Answer: A

Cheers,
Brent
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Re: The only contents of a container   [#permalink] 24 Nov 2019, 07:36
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