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A group of 3 integers is to be selected one after the other, [#permalink]

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24 Nov 2011, 11:45

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List L: 1, 2, 3, 4, 5, 6, 7, 8, 9, 10

A group of 3 integers is to be selected one after the other, and at random and without replacement, from list L above. What is the probability that the 3 integers selected are not 3 consecutive integers?

A. 3/5 B. 7/10 C. 3/4 D. 4/5 E. 14/15

One way to do this is to count all the consecutive combinations and divide that by 10C3, subtract the whole thing from 1. But is there a more systematic and efficient way to do this problem? kaplan books only provide brute force solutions without any formulas. thanks.

Re: The most efficient way to do this probability question [#permalink]

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24 Nov 2011, 12:36

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The method you suggested is the simplest. It's very easy to count the number of series 3 consecutive integers, especially since you don't need to do permutations. There are a total of 8 such series (1,2,3), (2,3,4)...(8,9,10).

A group of 3 integers is to be selected one after the other, and at random and without replacement, from list L above. What is the probability that the 3 integers selected are not 3 consecutive integers?

a - 3/5 b - 7/10 c - 3/4 d - 4/5 e - 14/15

One way to do this is to count all the consecutive combinations and divide that by 10C3, subtract the whole thing from 1. But is there a more systematic and efficient way to do this problem? kaplan books only provide brute force solutions without any formulas. thanks.

There is a little bit of ambiguity in this question. When you read it, you wonder whether the order of selection is important i.e. should the numbers be selected in a consecutive manner i.e. are they looking for the probability of selecting 2, then 3 and then 4 or is this selection made in any other order e.g. 3, then 2 and then 4 fine too. You need to guess from the answer options that the order of selection is not important. In that case, the solution provided above is the best and fastest method.

In case the order of selection is important, the required probability = (8/10)*(1/9)*(1/8) = 1/90 i.e. you can select any of the first 8 integers first. Now both the second and the third pick are defined e.g. if you select 4 on your first pick, you need to select 5 next (probability of that = 1/9 since there are total 9 numbers left) and then you need to select 6 (probability of that = 1/8 since there are total 8 numbers left)
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Re: The most efficient way to do this probability question [#permalink]

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25 Nov 2011, 05:39

As Karishma said , there is an ambiguity in the question. I guess one way to find the answer is to approach in both ways and check if only one answer is present among the answer choices.

If order of the consecutive numbers is not important , then answer would be 1- {8/10C3} or 1 - {8/120} or 14/15 If order of the consecutive numbers is important , then answer would be 1- {(8*3!)/10C3} or 1- {48/120} or 3/5.

In this case , both the answers are present in the answer choices ( ), so question needs to be little more clear. I don't think this kind of unclear question would ever come in actual GMAT .
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Wy do you restrict the first choice to 8 options?, the way I understand this problem you could pick any of the numbres 1 to 10 and then the next two would have constraints.

For example if my first pick is 10, my second 8 and my last pick is 9, wouldn't that set comply with being a set of consecutive numbers?

The point is 'does the order of picking numbers matter'. If only the end result is important i.e. you get three numbers which are consecutive, then the answer is different but if the order of selecting numbers is important too i.e. after 8, I must pick 9 and then 10, then the answer is different. The question would have been clearer if they had mentioned that three numbers are picked simultaneously. Then the order doesn't matter. When they say that the numbers are picked one after the other, it gets you thinking if the order of selection is important too. Anyway, the intention of the original question is what you suggested. I offered a different spin on it.

In fact, there are two different spins on the question.

When I restrict the first number to one of 1-8, I am assuming that they want the probability of 3 consecutive numbers picked in increasing order.

If you want to find the probability of 3 consecutive number picked in either increasing or decreasing order (say you pick 8, 9, 10 OR you pick 8, 7, 6 in that order), you do not need to have this restriction but you do need different cases. If you pick 1/2/9/10, the next two numbers can be chosen in only 1 way each. If you pick any other number, you can choose the next number in 2 ways and the third number in one way. Convert this logic into math and find out what you get.
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Re: The most efficient way to do this probability question [#permalink]

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27 Dec 2011, 05:40

I just thought of another way to figure out the problem:

1.) Select any number: P1 10/10 2.) Select any number next to the first one: P2 2/9 3.) Select and number next to the first or second chosen one: P3 2/8

P1 * P2 * P3 = 1/18

1 - 1/18 = 17/18 .... this is closest to E. The only problem is that I can also chose numbers 1 and 10 which don't have two neighboring numbers.

Taking this into account will make it too difficult and long-lasting (at least for me)

Re: The most efficient way to do this probability question [#permalink]

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07 Dec 2013, 01:14

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Re: A group of 3 integers is to be selected one after the other, [#permalink]

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12 Oct 2015, 03:40

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Re: A group of 3 integers is to be selected one after the other, [#permalink]

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09 Dec 2015, 20:08

I solved it using permutations: there are 10C3 number of ways to select 3 numbers, or 120 ways. I struggled to find the number of answers using probability and counting...but later, I just simply put 3 numbers in a "big" number, 3 numbers of course mean 3 consecutive numbers. Now, we have one "big" numbers and 7 other numbers. in how many ways can we select the big number? 8C1 or 8 ways. Thus, the probability of not being consecutive is (120-8)/120, which is 14/15

Determining that there were 120 ways to select 3 numbers from that group was the perfect first step. Next, you just have to figure out which combinations of 3 involve three consecutive integers. Thankfully, it's not that hard - and you can list them out:

123 234 345 456 567 678 789 89&10

Thus there are 8 ways (out of 120) that ARE three consecutive integers and the other 112/120 are NOT three consecutive integers.

A group of 3 integers is to be selected one after the other, [#permalink]

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10 Dec 2015, 20:03

EMPOWERgmatRichC wrote:

Hi mvictor,

Determining that there were 120 ways to select 3 numbers from that group was the perfect first step. Next, you just have to figure out which combinations of 3 involve three consecutive integers. Thankfully, it's not that hard - and you can list them out:

123 234 345 456 567 678 789 89&10

Thus there are 8 ways (out of 120) that ARE three consecutive integers and the other 112/120 are NOT three consecutive integers.

Thanks Rich for explanation. I usually have problems with combinatorics problems, especially when some restrictions are present. For example, in how many ways can we select 3 numbers so that the sum of these 3 numbers is not divisible by 2... Knowing that it is one of my greatest weaknesses, I sometimes overthink, and do not follow the "easy" step - actually counting the possibilities..

Many of these skills develop with time and practice. When these types of probability/permutation/combination questions show up on Test Day, the solution often 'leans' one way or the other, so it helps to think about which of two things would be easier to calculate:

1) What we want (what the question asks for) 2) What we don't want (the options that DON'T fit what the question asks for, so that we can subtract that number from the total to get the answer to the question).

Regardless, these types of questions are NOT worth a lot of points on Test Day. You shouldn't be spending too much time on these types of prompts if you're missing out on lots of points in the BIG Quant categories.

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