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Re: M2706
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11 May 2016, 23:04
sheetaldodani wrote: I think this is a poorquality question and the explanation isn't clear enough, please elaborate. I cannot understand the answer explanation of this one. i went to the forum and it is confusing there as well. In the egmat class the formula for probability is clearly stated as x/n where x is the number of ways in which the event is possible and n is the total possible outcomes. in this case, n would be no. of ways of selecting 5 integers out of the total number of integers multiplied with no. of ways by which they can be arranged. i.e 12C5 x 5P5 if this is true, then how does the original set i.e. K not matter? the questions states ' what is the probability that the numbers will be 'selected' in an ascending order' ... and not 'arranged' in an ascending order! hi, it is a good question.. when you have five items, they can be arranged in say x ways by whatever way you find it, BUT ONLY 1 out of entire ways is in ascending order... say lets go by your way.. total ways possible = kC5*5P5.. now how many ways will the 5 in ascending order, 1 way of each 5 selected = kC5.. prob = \(\frac{kC5}{kC5*5P5} = 1/5P5\).. so you see it doesn't matter WHAT k is, it depend on the number we are choosing out of k... Now even if you have 100 items and you are choosing 5 out of them, these 5 will be arranged in 5! ways and 1 out of these 5! will be in ascending order
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26 Jun 2016, 09:24
I think this is a poorquality question and the explanation isn't clear enough, please elaborate. We are selecting number "onebyone" so how is statement 2 alone is sufficient? Lets say we have 12 numbers in set k={1,2,3...,11,12} All of the following are ascending order selection 1,2,3,4,5 6,7,8,9,10 4,7,8,9,12 1,3,5,6,11 .....and many more possible cases. Please elaborate the solution provided is not at all clear. BUNUEL can you please help me understand this one?



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26 Jun 2016, 21:19
geek_mnnit wrote: I think this is a poorquality question and the explanation isn't clear enough, please elaborate. We are selecting number "onebyone" so how is statement 2 alone is sufficient? Lets say we have 12 numbers in set k={1,2,3...,11,12} All of the following are ascending order selection 1,2,3,4,5 6,7,8,9,10 4,7,8,9,12 1,3,5,6,11 .....and many more possible cases. Please elaborate the solution provided is not at all clear. BUNUEL can you please help me understand this one? Please got through the previous pages of discussion.
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06 Jul 2016, 04:49
Bunuel wrote: Official Solution:
We should understand the following two things: 1. The probability of selecting any \(n\) numbers from the set is the same. Why should any subset of \(n\) numbers have higher or lower probability of being selected than some other subset of \(n\) numbers? Probability doesn't favor any particular subset. 2. Now, consider that the subset selected is \(\{x_1, \ x_2, \ ..., \ x_n\}\), where \(x_1 \lt x_2 \lt ... \lt x_n\). We can select this subset of numbers in \(n!\) # of ways and out of these \(n!\) ways only one, namely \(\{x_1, \ x_2, \ ..., \ x_n\}\) will be in ascending order. So 1 out of n!. \(P=\frac{1}{n!}\). Hence, according to the above, the only thing we need to know to answer the question is the size of the subset (\(n\)) we are selecting from set \(A\).
Answer: B Dear Bunuel .. I have a concern here . I think the answer is D . As per the condition given the elements are selected one by one from the given set . So the probability varies as per this condition . For example , k=5 , n=3 , then the probability will be 1/5 * 1/4 *1/3 (each time select the least number from the remaining inputs) . So it should be 1/60 . I am not getting why you are taking n directly as a single input and calculating n! (one by one condition is not considered here) . Please let me know where I am going wrong ... Thanks in advance ..



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06 Jul 2016, 06:11
karnaidu wrote: Bunuel wrote: Official Solution:
We should understand the following two things: 1. The probability of selecting any \(n\) numbers from the set is the same. Why should any subset of \(n\) numbers have higher or lower probability of being selected than some other subset of \(n\) numbers? Probability doesn't favor any particular subset. 2. Now, consider that the subset selected is \(\{x_1, \ x_2, \ ..., \ x_n\}\), where \(x_1 \lt x_2 \lt ... \lt x_n\). We can select this subset of numbers in \(n!\) # of ways and out of these \(n!\) ways only one, namely \(\{x_1, \ x_2, \ ..., \ x_n\}\) will be in ascending order. So 1 out of n!. \(P=\frac{1}{n!}\). Hence, according to the above, the only thing we need to know to answer the question is the size of the subset (\(n\)) we are selecting from set \(A\).
Answer: B Dear Bunuel .. I have a concern here . I think the answer is D . As per the condition given the elements are selected one by one from the given set . So the probability varies as per this condition . For example , k=5 , n=3 , then the probability will be 1/5 * 1/4 *1/3 (each time select the least number from the remaining inputs) . So it should be 1/60 . I am not getting why you are taking n directly as a single input and calculating n! (one by one condition is not considered here) . Please let me know where I am going wrong ... Thanks in advance .. The first statement is not sufficient because the probability will be different for different values of n. Try for n=1, n=2, ... Please refer to the discussion on previous pages.
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Re: M2706
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06 Jul 2016, 19:45
Guess, everyone here is in doubt, cause they are thinking, numbers selected one by one, should be selected in ascending order. But question asks, numbers selected in set should be in ascending order.... irrespective of position they were selected. I personally think, bunuel soln is for special case k=n ... Will elaborate some other time.. But nice question



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Re: M2706
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07 Jul 2016, 20:01
sheetaldodani .. I believe it is a trick of words... In question they have already given n numbers are selected. So ideally no selection operation need to be done, but asking for the probability of a specific arrangement (i.e. when the numbers are in ascending order). So here we know there are totally n! cases and only one possible arrangement in the all the possible arrangements (n!). Hope it clears your doubt.



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20 Aug 2016, 03:51
I think this is a highquality question and I agree with explanation. Wow great question!



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20 Aug 2016, 18:27
Oh yes ... it does. I was clear with the earlier reply itself. Thank you so much



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23 Oct 2016, 12:11
Bunuel wrote: Official Solution:
We should understand the following two things: 1. The probability of selecting any \(n\) numbers from the set is the same. Why should any subset of \(n\) numbers have higher or lower probability of being selected than some other subset of \(n\) numbers? Probability doesn't favor any particular subset. 2. Now, consider that the subset selected is \(\{x_1, \ x_2, \ ..., \ x_n\}\), where \(x_1 \lt x_2 \lt ... \lt x_n\). We can select this subset of numbers in \(n!\) # of ways and out of these \(n!\) ways only one, namely \(\{x_1, \ x_2, \ ..., \ x_n\}\) will be in ascending order. So 1 out of n!. \(P=\frac{1}{n!}\). Hence, according to the above, the only thing we need to know to answer the question is the size of the subset (\(n\)) we are selecting from set \(A\).
Answer: B Hey Bunuel, where can I find more questions of this kind? Thanks in Advance



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23 Oct 2016, 21:51
amitn wrote: Bunuel wrote: Official Solution:
We should understand the following two things: 1. The probability of selecting any \(n\) numbers from the set is the same. Why should any subset of \(n\) numbers have higher or lower probability of being selected than some other subset of \(n\) numbers? Probability doesn't favor any particular subset. 2. Now, consider that the subset selected is \(\{x_1, \ x_2, \ ..., \ x_n\}\), where \(x_1 \lt x_2 \lt ... \lt x_n\). We can select this subset of numbers in \(n!\) # of ways and out of these \(n!\) ways only one, namely \(\{x_1, \ x_2, \ ..., \ x_n\}\) will be in ascending order. So 1 out of n!. \(P=\frac{1}{n!}\). Hence, according to the above, the only thing we need to know to answer the question is the size of the subset (\(n\)) we are selecting from set \(A\).
Answer: B Hey Bunuel, where can I find more questions of this kind? Thanks in Advance I wasn't able to find anything similar to this but you can check below post for more on probability: Probability Made Easy!Theory on probability problemsData Sufficiency Questions on ProbabilityProblem Solving Questions on ProbabilityTough Probability Questions
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Re: M2706
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23 Feb 2017, 00:02
first we will have to select n numbers from the set of k integers. Then only we can arrange in the ascending order as per the question. So, suppose there are 12 distinct integers in set k from which we need 5 integers (=n)........ then we can select by 12c5. This part is missing from the solution. As per this, only possible answer is (C). Kindly take this into account & post a firm explaination.



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M2706
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Updated on: 06 Apr 2017, 04:50
For those still confused, imagine two scenarios.
Imagine numbers 1 to 10. Make this a set. {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} If I told you to pick 3 numbers from this set, what is the probability of picking them in ascending order? Well, the number of possible arrangements of 3 numbers is 3!. Only one of those is ascending. So, only n matters and the probability is 1 out of 3! .
123 132 213 231 321 312
For scenario two, let's work backwards and imagine a similar question in a different light. Out of 5 numbers, what is the probability of arranging them in ascending order? Again, per the above it would be 1/5!. Think about this: this question is asking you to find how many arrangements of 5 numbers puts them in a sequence, i.e. ascending order. In any question that asks this, Set A is actually the set of all real, nonnegative integers AND k is INFINITY. So, you don't have to worry about the amount of numbers in Set A, but rather just the amount of numbers you are trying to arrange in a specific order, i.e. n.
Originally posted by mbadude2017 on 29 Mar 2017, 15:46.
Last edited by mbadude2017 on 06 Apr 2017, 04:50, edited 1 time in total.



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Re: M2706
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06 Apr 2017, 04:28
Hi..
This seems good question, however the explanation was not clear from beginning. Here are my views.
Statement1 : If we have 12 consecutive even number or any twelve 12 numbers does not matter. What matters here is : n, how many numbers we are going to select out of 12. Based on different number of n , the probability will be different. Ex: for n = 2 1,2; 1,3;1,4; so on... 2,3; 2,4; 2,5; so on.. so on.. for other numbers same goes for n=3 and probability will decrease with less combination of ascending orders. Not Sufficient
Statement2: If n is fixed then we always have fixed probability w.r.t. n for n=2 ; 1/factorial(2) : 1/2 for n=3 ; 1/factorial(3) : 1/6
but the problem is K we don't know. same as statement 1. so it should be also insufficient.
However by adding Stat1 & 2 , it should be sufficient , hence "C" should be answer.
Any Comments ?



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06 Apr 2017, 04:41
MountainGMAT wrote: Hi..
This seems good question, however the explanation was not clear from beginning. Here are my views.
Statement1 : If we have 12 consecutive even number or any twelve 12 numbers does not matter. What matters here is : n, how many numbers we are going to select out of 12. Based on different number of n , the probability will be different. Ex: for n = 2 1,2; 1,3;1,4; so on... 2,3; 2,4; 2,5; so on.. so on.. for other numbers same goes for n=3 and probability will decrease with less combination of ascending orders. Not Sufficient
Statement2: If n is fixed then we always have fixed probability w.r.t. n for n=2 ; 1/factorial(2) : 1/2 for n=3 ; 1/factorial(3) : 1/6
but the problem is K we don't know. same as statement 1. so it should be also insufficient.
However by adding Stat1 & 2 , it should be sufficient , hence "C" should be answer.
Any Comments ? The answer is B not C. For (2) we are given that n =5, so the probability will be 1/5! regardless of k.
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24 Apr 2017, 18:32
I got this question wrong, but I guess I get the idea now. Here is what I understood (Pardon the formatting) : Numerator = Favourable outcomes. Selection and Arrangement, because order is important. Number of ways to select n things out of k things = KCn Number of ways to arrange the selection in ascending order = 1 way Numerator = Selection and Arrangement = KCn * 1 Denominator = Total Outcomes. Selection and arrangement Selection = Number of ways to select n things out of K things = KCn Arrangement = Number of ways to arrange n things = n! Probability = Favourable outcomes / Total Outcomes. Probability = (KCn *1) / (KCn * n!) = 1/n! Hence Option B



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Bunuel wrote: Set \(A\) consists of \(k\) distinct numbers. If \(n\) numbers are selected from the set onebyone, where \(n \le k\), what is the probability that numbers will be selected in ascending order?
(1) Set \(A\) consists of 12 even consecutive integers.
(2) \(n=5\). I understood statement 2 very easily. Since you are pulling 5 distinct numbers, it doesn't matter how big the pool is, since all numbers are distinct you can figure out the probability of whether or not the numbers you pull all ascending or not. For statement 1, we don't know how many that we are going to pull. But does that really make it impossible to find the probability of pulling an ascending series of numbers? Since we know k=12, we know that 0<=n<=12. Couldn't we find the probability like this: ProbOfAscendingWithN=0 OR ProbOfAscendingWithN=1 OR ... OR ProbOfAscendingWithN=12? Basically find 13 different probabilities and combine?



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12 Jun 2017, 11:06
I think this is a poorquality question and the explanation isn't clear enough, please elaborate. I have a doubt on this one (and many others as well), but I guess if we are to select a subset of n numbers from a set with k distinct numbers we should have k!/(kn)! ways to select and not just n!. If we are to choose 3 elements from a set of 5 elements we have 5*4*3.
In this case we would need both, k and n values, no?!
Thank you for further explanations! (Y)



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12 Jun 2017, 12:20
leandrodijon wrote: I think this is a poorquality question and the explanation isn't clear enough, please elaborate. I have a doubt on this one (and many others as well), but I guess if we are to select a subset of n numbers from a set with k distinct numbers we should have k!/(kn)! ways to select and not just n!. If we are to choose 3 elements from a set of 5 elements we have 5*4*3.
In this case we would need both, k and n values, no?!
Thank you for further explanations! (Y) First of all, this is a hard question so you should be comfortable with combinations and probability to solve it. Next, if we are choosing 5 numbers, the probability of choosing them in ascending order is 1/5!. The number of permutations of 5 different number, the number of ways we can choose 5 different number, say 1, 2, 3, 4 and 5 is 5! = 120: {1, 2, 3, 4, 5} {1, 2, 3, 5, 4} {1, 2, 4, 3, 5} {1, 2, 4, 5, 3} {1, 2, 5, 3, 4} {1, 2, 5, 4, 3} {1, 3, 2, 4, 5} {1, 3, 2, 5, 4} {1, 3, 4, 2, 5} {1, 3, 4, 5, 2} {1, 3, 5, 2, 4} {1, 3, 5, 4, 2} {1, 4, 2, 3, 5} {1, 4, 2, 5, 3} {1, 4, 3, 2, 5} {1, 4, 3, 5, 2} {1, 4, 5, 2, 3} {1, 4, 5, 3, 2} {1, 5, 2, 3, 4} {1, 5, 2, 4, 3} {1, 5, 3, 2, 4} {1, 5, 3, 4, 2} {1, 5, 4, 2, 3} {1, 5, 4, 3, 2} {2, 1, 3, 4, 5} {2, 1, 3, 5, 4} {2, 1, 4, 3, 5} {2, 1, 4, 5, 3} {2, 1, 5, 3, 4} {2, 1, 5, 4, 3} {2, 3, 1, 4, 5} {2, 3, 1, 5, 4} {2, 3, 4, 1, 5} {2, 3, 4, 5, 1} {2, 3, 5, 1, 4} {2, 3, 5, 4, 1} {2, 4, 1, 3, 5} {2, 4, 1, 5, 3} {2, 4, 3, 1, 5} {2, 4, 3, 5, 1} {2, 4, 5, 1, 3} {2, 4, 5, 3, 1} {2, 5, 1, 3, 4} {2, 5, 1, 4, 3} {2, 5, 3, 1, 4} {2, 5, 3, 4, 1} {2, 5, 4, 1, 3} {2, 5, 4, 3, 1} {3, 1, 2, 4, 5} {3, 1, 2, 5, 4} {3, 1, 4, 2, 5} {3, 1, 4, 5, 2} {3, 1, 5, 2, 4} {3, 1, 5, 4, 2} {3, 2, 1, 4, 5} {3, 2, 1, 5, 4} {3, 2, 4, 1, 5} {3, 2, 4, 5, 1} {3, 2, 5, 1, 4} {3, 2, 5, 4, 1} {3, 4, 1, 2, 5} {3, 4, 1, 5, 2} {3, 4, 2, 1, 5} {3, 4, 2, 5, 1} {3, 4, 5, 1, 2} {3, 4, 5, 2, 1} {3, 5, 1, 2, 4} {3, 5, 1, 4, 2} {3, 5, 2, 1, 4} {3, 5, 2, 4, 1} {3, 5, 4, 1, 2} {3, 5, 4, 2, 1} {4, 1, 2, 3, 5} {4, 1, 2, 5, 3} {4, 1, 3, 2, 5} {4, 1, 3, 5, 2} {4, 1, 5, 2, 3} {4, 1, 5, 3, 2} {4, 2, 1, 3, 5} {4, 2, 1, 5, 3} {4, 2, 3, 1, 5} {4, 2, 3, 5, 1} {4, 2, 5, 1, 3} {4, 2, 5, 3, 1} {4, 3, 1, 2, 5} {4, 3, 1, 5, 2} {4, 3, 2, 1, 5} {4, 3, 2, 5, 1} {4, 3, 5, 1, 2} {4, 3, 5, 2, 1} {4, 5, 1, 2, 3} {4, 5, 1, 3, 2} {4, 5, 2, 1, 3} {4, 5, 2, 3, 1} {4, 5, 3, 1, 2} {4, 5, 3, 2, 1} {5, 1, 2, 3, 4} {5, 1, 2, 4, 3} {5, 1, 3, 2, 4} {5, 1, 3, 4, 2} {5, 1, 4, 2, 3} {5, 1, 4, 3, 2} {5, 2, 1, 3, 4} {5, 2, 1, 4, 3} {5, 2, 3, 1, 4} {5, 2, 3, 4, 1} {5, 2, 4, 1, 3} {5, 2, 4, 3, 1} {5, 3, 1, 2, 4} {5, 3, 1, 4, 2} {5, 3, 2, 1, 4} {5, 3, 2, 4, 1} {5, 3, 4, 1, 2} {5, 3, 4, 2, 1} {5, 4, 1, 2, 3} {5, 4, 1, 3, 2} {5, 4, 2, 1, 3} {5, 4, 2, 3, 1} {5, 4, 3, 1, 2} {5, 4, 3, 2, 1} (total: 120) Only one of the total of 120 arrangements is in ascending order, namely the first one: {1, 2, 3, 4, 5}. So, the probability is 1/120.
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