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The next set of tough and tricky DS questions. I'll post OA's with detailed explanations after some discussion. Please, post your solutions along with the answers. Good luck!
1. Jules and Jim both invested certain amount of money in bond M for one year, which pays for 12% simple interest annually. If no other investment were made, then Jules initial investment in bond M was how many dollars more than Jim's investment in bond M. (1) In one year Jules earned $24 more than Jim from bond M. (2) If the interest were 20% then in one year Jules would have earned $40 more than Jim from bond M.
4. Of the 58 patients of Vertigo Hospital, 45 have arachnophobia. How many of the patients have acrophobia? (1) The number of patients of Vertigo Hospital who have both arachnophobia and acrophobia is the same as the number of patients who have neither arachnophobia nor acrophobia. (2) 32 patients of Vertigo Hospital have arachnophobia but not acrophobia.
5. If at least one astronaut do NOT listen to Bach at Solaris space station, then how many of 35 astronauts at Solaris space station listen to Bach? (1) Of the astronauts who do NOT listen to Bach 56% are male. (2) Of the astronauts who listen to Bach 70% are female.
7. Set A consists of k distinct numbers. If n numbers are selected from the set one-by-one, where n<=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.
9. The product of three distinct positive integers is equal to the square of the largest of the three numbers, what is the product of the two smaller numbers? (1) The average (arithmetic mean) of the three numbers is 34/3. (2) The largest number of the three distinct numbers is 24.
10. There is at least one viper and at least one cobra in Pandora's box. How many cobras are there? (1) There are total 99 snakes in Pandora's box. (2) From any two snakes from Pandora's box at least one is a viper.
11. Alice has $15, which is enough to buy 11 muffins and 7 brownies, is $45 enough to buy 27 muffins and 27 brownies? (1) $15 is enough to buy 7 muffins and 11 brownies. (2) $15 is enough to buy 10 muffins and 8 brownies.
13. Buster leaves the trailer at noon and walks towards the studio at a constant rate of B miles per hour. 20 minutes later, Charlie leaves the same studio and walks towards the same trailer at a constant rate of C miles per hour along the same route as Buster. Will Buster be closer to the trailer than to the studio when he passes Charlie? (1) Charlie gets to the trailer in 55 minutes. (2) Buster gets to the studio at the same time as Charlie gets to the trailer.
Re: Devil's Dozen!!! [#permalink]
23 Mar 2012, 02:53
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10. There is at least one viper and at least one cobra in Pandora's box. How many cobras are there?
Quite tricky.
(1) There are total 99 snakes in Pandora's box. Clearly insufficient.
(2) From any two snakes from Pandora's box at least one is a viper. Since from ANY two snakes one is a viper then there can not be 2 (or more) cobras and since there is at least one cobra then there must be exactly one cobra in the box. Sufficient.
Re: Devil's Dozen!!! [#permalink]
23 Mar 2012, 02:55
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11. Alice has $15, which is enough to buy 11 muffins and 7 brownies, is $45 enough to buy 27 muffins and 27 brownies?
700+ question.
Given: \(11m+7b\leq{15}\), where \(m\) and \(b\) are prices of one muffin and one brownie respectively. Question: is \(27m+27b\leq{45}\)? --> \(9m+9b\leq{15}\). Question basically asks whether we can substitute 2 muffins with 2 brownies.
Now if \(m>b\) we can easily substitute 2 muffins with 2 brownies (since \(2m\) will be more than \(2b\)). But if \(m<b\) we won't know this for sure.
But consider the case when we are told that we can substitute 3 muffins with 3 brownies. In both cases (\(m>b\) or \(m<b\)) it would mean that we can substitute 2 (so less than 3) muffins with 2 brownies, but again we won't be sure whether we can substitute 4 (so more than 3) muffins with 4 brownies.
(1) $15 is enough to buy 7 muffins and 11 brownies --> \(7m+11b\leq{15}\): we can substitute 4 muffins with 4 brownies, so according to above we can surely substitute 2 muffins with 2 brownies. Sufficient. (1) $15 is enough to buy 10 muffins and 8 brownies --> \(10m+8b\leq{15}\): we can substitute 1 muffin with 1 brownie, so according to above this is does not ensure that we can substitute 2 muffins with 2 brownies. Not sufficient.
Re: Devil's Dozen!!! [#permalink]
23 Mar 2012, 02:57
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13. Buster leaves the trailer at noon and walks towards the studio at a constant rate of B miles per hour. 20 minutes later, Charlie leaves the same studio and walks towards the same trailer at a constant rate of C miles per hour along the same route as Buster. Will Buster be closer to the trailer than to the studio when he passes Charlie?
(1) Charlie gets to the trailer in 55 minutes. No info about Buster. Not sufficient.
(2) Buster gets to the studio at the same time as Charlie gets to the trailer --> Charlie needed 20 minutes less than Buster to cover the same distance, which means that the rate of Charlie is higher than that of Buster. Since after they pass each other they need the same time to get to their respective destinations (they get at the same time to their respective destinations) then Buster had less distance to cover ahead (at lower rate) than he had already covered (which would be covered by Charlie at higher rate). Sufficient.
Re: Devil's Dozen!!! [#permalink]
23 Mar 2012, 04:25
Bunuel wrote:
11. Alice has $15, which is enough to buy 11 muffins and 7 brownies, is $45 enough to buy 27 muffins and 27 brownies?
700+ question.
Given: \(11m+7b\leq{15}\), where \(m\) and \(b\) are prices of one muffin and one brownie respectively. Question: is \(27m+27b\leq{45}\)? --> \(9m+9b\leq{15}\). Question basically asks whether we can substitute 2 muffins with 2 brownies.
Now if \(m>b\) we can easily substitute 2 muffins with 2 brownies (since \(2m\) will be more than \(2b\)). But if \(m<b\) we won't know this for sure.
But consider the case when we are told that we can substitute 3 muffins with 3 brownies. In both cases (\(m>b\) or \(m<b\)) it would mean that we can substitute 2 (so less than 3) muffins with 2 brownies, but again we won't be sure whether we can substitute 4 (so more than 3) muffins with 4 brownies.
(1) $15 is enough to buy 7 muffins and 11 brownies --> \(7m+11b\leq{15}\): we can substitute 4 muffins with 4 brownies, so according to above we can surely substitute 2 muffins with 2 brownies. Sufficient. (1) $15 is enough to buy 10 muffins and 8 brownies --> \(10m+8b\leq{15}\): we can substitute 1 muffin with 1 brownie, so according to above this is does not ensure that we can substitute 2 muffins with 2 brownies. Not sufficient.
Answer: A.
Thanks for the solutions Bunuel. How to get this qn solved in this way. I mean, i tried solving this qn in the usual algebraic way, as the qn sounded more that way n i got D. i'm not getting why its wrong that way. can you please help in this. Thanks for explaining.
Re: Devil's Dozen!!! [#permalink]
23 Mar 2012, 04:28
Expert's post
priyavenugopal wrote:
Bunuel wrote:
11. Alice has $15, which is enough to buy 11 muffins and 7 brownies, is $45 enough to buy 27 muffins and 27 brownies?
700+ question.
Given: \(11m+7b\leq{15}\), where \(m\) and \(b\) are prices of one muffin and one brownie respectively. Question: is \(27m+27b\leq{45}\)? --> \(9m+9b\leq{15}\). Question basically asks whether we can substitute 2 muffins with 2 brownies.
Now if \(m>b\) we can easily substitute 2 muffins with 2 brownies (since \(2m\) will be more than \(2b\)). But if \(m<b\) we won't know this for sure.
But consider the case when we are told that we can substitute 3 muffins with 3 brownies. In both cases (\(m>b\) or \(m<b\)) it would mean that we can substitute 2 (so less than 3) muffins with 2 brownies, but again we won't be sure whether we can substitute 4 (so more than 3) muffins with 4 brownies.
(1) $15 is enough to buy 7 muffins and 11 brownies --> \(7m+11b\leq{15}\): we can substitute 4 muffins with 4 brownies, so according to above we can surely substitute 2 muffins with 2 brownies. Sufficient. (1) $15 is enough to buy 10 muffins and 8 brownies --> \(10m+8b\leq{15}\): we can substitute 1 muffin with 1 brownie, so according to above this is does not ensure that we can substitute 2 muffins with 2 brownies. Not sufficient.
Answer: A.
Thanks for the solutions Bunuel. How to get this qn solved in this way. I mean, i tried solving this qn in the usual algebraic way, as the qn sounded more that way n i got D. i'm not getting why its wrong that way. can you please help in this. Thanks for explaining.
Can you please show your algebraic solution to point out possible flaws in it? _________________
Re: Devil's Dozen!!! [#permalink]
23 Mar 2012, 05:28
Bunuel wrote:
priyavenugopal wrote:
Bunuel wrote:
11. Alice has $15, which is enough to buy 11 muffins and 7 brownies, is $45 enough to buy 27 muffins and 27 brownies?
700+ question.
Given: \(11m+7b\leq{15}\), where \(m\) and \(b\) are prices of one muffin and one brownie respectively. Question: is \(27m+27b\leq{45}\)? --> \(9m+9b\leq{15}\). Question basically asks whether we can substitute 2 muffins with 2 brownies.
Now if \(m>b\) we can easily substitute 2 muffins with 2 brownies (since \(2m\) will be more than \(2b\)). But if \(m<b\) we won't know this for sure.
But consider the case when we are told that we can substitute 3 muffins with 3 brownies. In both cases (\(m>b\) or \(m<b\)) it would mean that we can substitute 2 (so less than 3) muffins with 2 brownies, but again we won't be sure whether we can substitute 4 (so more than 3) muffins with 4 brownies.
(1) $15 is enough to buy 7 muffins and 11 brownies --> \(7m+11b\leq{15}\): we can substitute 4 muffins with 4 brownies, so according to above we can surely substitute 2 muffins with 2 brownies. Sufficient. (1) $15 is enough to buy 10 muffins and 8 brownies --> \(10m+8b\leq{15}\): we can substitute 1 muffin with 1 brownie, so according to above this is does not ensure that we can substitute 2 muffins with 2 brownies. Not sufficient.
Answer: A.
Thanks for the solutions Bunuel. How to get this qn solved in this way. I mean, i tried solving this qn in the usual algebraic way, as the qn sounded more that way n i got D. i'm not getting why its wrong that way. can you please help in this. Thanks for explaining.
Can you please show your algebraic solution to point out possible flaws in it?
Sure. Thanks. I just tried it in the vry usual algebraic eqn solving way.
Given, From qn stem - 11m + 7b = 15 (i took the 'enough' word as 'equal to'..as the 'enough' keyword can be rephrased as '<=', i thought that,considering the max limit '=' would do in such price lmt case..not sure, whthr thrs any trap thr..) From (1) stmt - 7m + 11b = 15 From (2) stmt - 10m + 8b = 15 Solving qn stem n stmt 1 eqns, got m and b as 5/6 (didnt get an int val thou..not sure whether thrs a trap) Solving qn stem n stmt 2 eqns, got m and b as 5/6
Having m and b values, we can find whthr 45$ is enough to buy 27m + 27b or not. (i.e. applying m n b in eqn 27m + 27b = 45 => 2 * 27 * 5/6 = 45). So, arrived at D.
Re: Devil's Dozen!!! [#permalink]
23 Mar 2012, 05:36
Expert's post
priyavenugopal wrote:
Sure. Thanks. I just tried it in the vry usual algebraic eqn solving way.
Given, From qn stem - 11m + 7b = 15 (i took the 'enough' word as 'equal to'..as the 'enough' keyword can be rephrased as '<=', i thought that,considering the max limit '=' would do in such price lmt case..not sure, whthr thrs any trap thr..) From (1) stmt - 7m + 11b = 15 From (2) stmt - 10m + 8b = 15 Solving qn stem n stmt 1 eqns, got m and b as 5/6 (didnt get an int val thou..not sure whether thrs a trap) Solving qn stem n stmt 2 eqns, got m and b as 5/6
Having m and b values, we can find whthr 45$ is enough to buy 27m + 27b or not. (i.e. applying m n b in eqn 27m + 27b = 45 => 2 * 27 * 5/6 = 45). So, arrived at D.
Not even to analyze the rest of it, I must say that there is a huge difference between \(11m+7b\leq{15}\) and \(11m+7={15}\). You can not just write = sing instead of <= just because it's more convenient and "enough" should be translated as <= only. _________________
Re: Devil's Dozen!!! [#permalink]
23 Mar 2012, 09:18
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Bunuel wrote:
2. If n is a positive integer and p is a prime number, is p a factor of n!?
(1) p is a factor of (n+2)!-n! --> if \(n=2\) then \((n+2)!-n!=22\) and for \(p=2\) then answer will be YES but for \(p=11\) the answer will be NO. Not sufficient.
(2) p is a factor of (n+2)!/n! --> \(\frac{(n+2)!}{n!}=(n+1)(n+2)\) --> if \(n=2\) then \((n+1)(n+2)=12\) and for \(p=2\) then answer will be YES but for \(p=3\) the answer will be NO. Not sufficient.
(1)+(2) \((n+2)!-n!=n!((n+1)(n+2)-1)\). Now, \((n+1)(n+2)-1\) and \((n+1)(n+2)\) are consecutive integers. Two consecutive integers are co-prime, which means that they don't share ANY common factor but 1. For example 20 and 21 are consecutive integers, thus only common factor they share is 1. So, as from (2) \(p\) is a factor of \((n+1)(n+2)\) then it can not be a factor of \((n+1)(n+2)-1\), thus in order \(p\) to be a factor of \(n!*((n+1)(n+2)-1)\), from (1), then it should be a factor of the first multiple of this expression: \(n!\). Sufficient.
Answer: C.
I didnot understand why 1 is not sufficient, can you elaborate on that ?
what i did is - I took n! common * coprime number
So if we have n! everytime the factors for n! and n!* coprime should always be same.. What am i missing?
Will Vote for A
Given n = +ve Int P = Prime
(A)
p is a factor of (n+2)!-n!
Let n = 4
= 6! - 4! = 4![(6*5) - 1] = 4! (29)
therefore n! = 4! will have sufficient factors as 4! (29)
Re: Devil's Dozen!!! [#permalink]
23 Mar 2012, 17:39
Bunuel wrote:
7. Set A consists of k distinct numbers. If n numbers are selected from the set one-by-one, where n<=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.
We should understand 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 \(x_1<x_2<...<x_n\). 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.
I am not sure if I understand correctly.
There are a total of k integers. So probability of selecting first integer 1/k, now probability of second integer out of k-1 integer is 1/(k-1) and so on... so selecting n integers would be 1/(k*(k-1)*(k-2)*...(k-(n-1)) so we need both k & n. Am I missing something here?
Re: Devil's Dozen!!! [#permalink]
23 Mar 2012, 17:44
Bunuel wrote:
10. There is at least one viper and at least one cobra in a Pandora's box. How many cobras are there?
Quite tricky.
(1) There are total 99 snakes in Pandora's box. Clearly insufficient.
(2) From any two snakes from the a Pandora's box at least one is a viper. Since from ANY two snakes one is a viper then there can not be 2 (or more) cobras and since there is at least one cobra then there must be exactly one cobra in the box. Sufficient.
Answer: B.
How do we know that there are no other varieties of snakes other than cobra and viper. This is precisely why I chose E.
Re: Devil's Dozen!!! [#permalink]
24 Mar 2012, 02:16
Expert's post
kuttingchai wrote:
Bunuel wrote:
2. If n is a positive integer and p is a prime number, is p a factor of n!?
(1) p is a factor of (n+2)!-n! --> if \(n=2\) then \((n+2)!-n!=22\) and for \(p=2\) then answer will be YES but for \(p=11\) the answer will be NO. Not sufficient.
(2) p is a factor of (n+2)!/n! --> \(\frac{(n+2)!}{n!}=(n+1)(n+2)\) --> if \(n=2\) then \((n+1)(n+2)=12\) and for \(p=2\) then answer will be YES but for \(p=3\) the answer will be NO. Not sufficient.
(1)+(2) \((n+2)!-n!=n!((n+1)(n+2)-1)\). Now, \((n+1)(n+2)-1\) and \((n+1)(n+2)\) are consecutive integers. Two consecutive integers are co-prime, which means that they don't share ANY common factor but 1. For example 20 and 21 are consecutive integers, thus only common factor they share is 1. So, as from (2) \(p\) is a factor of \((n+1)(n+2)\) then it can not be a factor of \((n+1)(n+2)-1\), thus in order \(p\) to be a factor of \(n!*((n+1)(n+2)-1)\), from (1), then it should be a factor of the first multiple of this expression: \(n!\). Sufficient.
Answer: C.
I didnot understand why 1 is not sufficient, can you elaborate on that ?
There are two examples given which give two different answers to the question whether p is a factor of n!.
(1) p is a factor of (n+2)!-n!
If \(n=2\) and \(p=2\) (notice that in this case \((n+2)!-n!=22\) and \(p=2\) is a factor of 22), then since \(p=2\) is a factor of \(n!=2!\) the answer to the question is YES;
If \(n=2\) and \(p=11\) (notice that in this case \((n+2)!-n!=22\) and \(p=11\) is a factor of 22), then since \(p=11\) is NOT a factor of \(n!=2!\) the answer to the question is NO.
Two different answers, hence not sufficient. _________________
Re: Devil's Dozen!!! [#permalink]
24 Mar 2012, 02:22
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mithun2vrs wrote:
Bunuel wrote:
7. Set A consists of k distinct numbers. If n numbers are selected from the set one-by-one, where n<=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.
We should understand 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 \(x_1<x_2<...<x_n\). 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.
I am not sure if I understand correctly.
There are a total of k integers. So probability of selecting first integer 1/k, now probability of second integer out of k-1 integer is 1/(k-1) and so on... so selecting n integers would be 1/(k*(k-1)*(k-2)*...(k-(n-1)) so we need both k & n. Am I missing something here?
We are not interested in the probability of selecting first, second, etc numbers from the set. We are interested in the probability that numbers selected will be in ascending order.
Now, any group of n numbers is equally likely to be selected and these n numbers can be selected in n! ways, out of which only 1 will be in ascending order so the probability is 1/n!.
Re: Devil's Dozen!!! [#permalink]
24 Mar 2012, 02:35
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Expert's post
mithun2vrs wrote:
Bunuel wrote:
10. There is at least one viper and at least one cobra in a Pandora's box. How many cobras are there?
Quite tricky.
(1) There are total 99 snakes in Pandora's box. Clearly insufficient.
(2) From any two snakes from Pandora's box at least one is a viper. Since from ANY two snakes one is a viper then there can not be 2 (or more) cobras and since there is at least one cobra then there must be exactly one cobra in the box. Sufficient.
Answer: B.
How do we know that there are no other varieties of snakes other than cobra and viper. This is precisely why I chose E.
Answer to the question is B, not E.
If there is some other snake, then group of {other, cobra} will be possible and the statement (2) will be violated (basically the possibility of other variety of snake is ruled out by the same logic as the possibility of second cobra).
Re: Devil's Dozen!!! [#permalink]
24 Mar 2012, 05:43
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Bunuel wrote:
mithun2vrs wrote:
Bunuel wrote:
7. Set A consists of k distinct numbers. If n numbers are selected from the set one-by-one, where n<=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.
We should understand 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 \(x_1<x_2<...<x_n\). 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.
I am not sure if I understand correctly.
There are a total of k integers. So probability of selecting first integer 1/k, now probability of second integer out of k-1 integer is 1/(k-1) and so on... so selecting n integers would be 1/(k*(k-1)*(k-2)*...(k-(n-1)) so we need both k & n. Am I missing something here?
We are not interested in the probability of selecting first, second, etc numbers from the set. We are interested in the probability that numbers selected will be in ascending order.
Now, any group of n numbers is equally likely to be selected and these n numbers can be selected in n! ways, out of which only 1 will be in ascending order so the probability is 1/n!.
Hope it's clear.
I think I will still disagree, it is a classic case of choosing numbers after replacement or without replacement. You interpret as earlier option and I later. Your thots pls
Re: Devil's Dozen!!! [#permalink]
24 Mar 2012, 05:51
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mithun2vrs wrote:
I think I will still disagree, it is a classic case of choosing numbers after replacement or without replacement. You interpret as earlier option and I later. Your thots pls
I think you misinterpret the question. We are not talking about the replacement at all, we are talking about the probability of selecting some group in certain order. Just try to calculate the probability of choosing 5 numbers out of 12 distinct numbers in ascending order and compare it to the probability of choosing 5 numbers out of 24 distinct numbers in ascending order. _________________
Re: Devil's Dozen!!! [#permalink]
24 Mar 2012, 07:41
Bunuel wrote:
mithun2vrs wrote:
I think I will still disagree, it is a classic case of choosing numbers after replacement or without replacement. You interpret as earlier option and I later. Your thots pls
I think you misinterpret the question. We are not talking about the replacement at all, we are talking about the probability of selecting some group in certain order. Just try to calculate the probability of choosing 5 numbers out of 12 distinct numbers in ascending order and compare it to the probability of choosing 5 numbers out of 24 distinct numbers in ascending order.
Sorry to keep on with the problem,
Probability is defined as No of fav outcomes / No of Total possible outcomes
With your example, in both cases No of favorable outcomes is 1 and in 1st case No of total possible outcomes is 12c5 and in the second case No of total possible outcomes is 24c5. Pls correct me where I am going wrong.
Re: Devil's Dozen!!! [#permalink]
24 Mar 2012, 13:52
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Expert's post
mithun2vrs wrote:
Bunuel wrote:
mithun2vrs wrote:
I think I will still disagree, it is a classic case of choosing numbers after replacement or without replacement. You interpret as earlier option and I later. Your thots pls
I think you misinterpret the question. We are not talking about the replacement at all, we are talking about the probability of selecting some group in certain order. Just try to calculate the probability of choosing 5 numbers out of 12 distinct numbers in ascending order and compare it to the probability of choosing 5 numbers out of 24 distinct numbers in ascending order.
Sorry to keep on with the problem,
Probability is defined as No of fav outcomes / No of Total possible outcomes
With your example, in both cases No of favorable outcomes is 1 and in 1st case No of total possible outcomes is 12c5 and in the second case No of total possible outcomes is 24c5. Pls correct me where I am going wrong.
Again: you count the probability of selecting 5 numbers out of 12 or 24, whereas the question asks about the probability of selecting some group of numbers in a certain ORDER. You just interpret the question in a wrong way. _________________
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