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According to a recent student poll, 5/7 out of 21 members of [#permalink]
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Updated on: 26 Jul 2013, 13:31
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According to a recent student poll, 5/7 out of 21 members of the finance club are interested in a career in investment banking. If two students are chosen at random, what is the probability that at least one of them is interested in investment banking? A. 1/14 B. 4/49 C. 2/7 D. 45/49 E. 13/14
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Originally posted by marcodonzelli on 05 Feb 2008, 10:40.
Last edited by Bunuel on 26 Jul 2013, 13:31, edited 1 time in total.
Edited the question and moved to PS forum.



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Re: complementary events  probabilities [#permalink]
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05 Feb 2008, 11:04
marcodonzelli wrote: according to a recent student poll, 5/7 of the 21 members of the finance club are interested in a career in investment banking. if 2 students are chosen at random, what is the probability that at least one of them is interested in investment banking?
1/14 4/49 2/7 45/49 13/14 1none none = 2/7 * 1/4 = 1/14 11/14 = 1/13
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Re: complementary events  probabilities [#permalink]
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05 Feb 2008, 11:37
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marcodonzelli wrote: according to a recent student poll, 5/7 of the 21 members of the finance club are interested in a career in investment banking. if 2 students are chosen at random, what is the probability that at least one of them is interested in investment banking?
1/14 4/49 2/7 45/49 13/14 15 students are interested, 6 are not interested Prob = 1  6C2/21C2 = 1  (6*5/(21*20))=1  1/14 = 13/14 > E



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Re: complementary events  probabilities [#permalink]
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05 Feb 2008, 11:44
Just a quick question, maratikus how do I know have to use combination in probability,
I mean my logic was,
15 interested, 6 not interested.
(1/15)(1/6) + (1/15)(1/6) + (2/15)
i.e. selected one from each group or select 2 from the interested group. This will insure we have atleast one from the interested.
Where am I going wrong?
Thanks Jack



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Re: complementary events  probabilities [#permalink]
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05 Feb 2008, 11:45
bmwhype2 wrote: marcodonzelli wrote: according to a recent student poll, 5/7 of the 21 members of the finance club are interested in a career in investment banking. if 2 students are chosen at random, what is the probability that at least one of them is interested in investment banking?
1/14 4/49 2/7 45/49 13/14 1none none = 2/7 * 1/4 = 1/14 11/14 = 1/13 I get the 1none and the 2/7 but where does the 1/4 come from?



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Re: complementary events  probabilities [#permalink]
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05 Feb 2008, 11:54
lsuguy7 wrote: bmwhype2 wrote: marcodonzelli wrote: according to a recent student poll, 5/7 of the 21 members of the finance club are interested in a career in investment banking. if 2 students are chosen at random, what is the probability that at least one of them is interested in investment banking?
1/14 4/49 2/7 45/49 13/14 1none none = 2/7 * 1/4 = 1/14 11/14 = 1/13 I get the 1none and the 2/7 but where does the 1/4 come from? 6 students are not interested in investments, 21 total > the first prob 6/21=2/7, 5 students are not interested in investments, 20 total > the second 5/20 = 1/4.



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Re: complementary events  probabilities [#permalink]
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05 Feb 2008, 12:53
bmwhype2 wrote: marcodonzelli wrote: according to a recent student poll, 5/7 of the 21 members of the finance club are interested in a career in investment banking. if 2 students are chosen at random, what is the probability that at least one of them is interested in investment banking?
1/14 4/49 2/7 45/49 13/14 1none none = 2/7 * 1/4 = 1/14 11/14 = 1/13 whoops. subtracted wrong...
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Re: complementary events  probabilities [#permalink]
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05 Feb 2008, 13:04
jackychamp wrote: Just a quick question, maratikus how do I know have to use combination in probability,
I mean my logic was,
15 interested, 6 not interested.
(1/15)(1/6) + (1/15)(1/6) + (2/15)
i.e. selected one from each group or select 2 from the interested group. This will insure we have atleast one from the interested.
Where am I going wrong?
Thanks Jack the first person we choose is either interested 15/21 (works) or not interested 6/21 > then the second person we choose has to be interested 15/20 15/21 + (6/21)*(15/20) = 5/7 + (2/7)*(3/4)=5/7+ 6/28 = 10/14 + 3/14 = 13/14 > E is that what you were considering?



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Re: complementary events  probabilities [#permalink]
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05 Feb 2008, 13:31
Thanks maratikus.
I was counting the probability of atleast one twice.
i.e.
prob of both + prob of one + prob of one which is wrong.
Thanks for the explanation.
Jack



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Re: complementary events  probabilities [#permalink]
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05 Feb 2008, 19:38
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marcodonzelli wrote: according to a recent student poll, 5/7 of the 21 members of the finance club are interested in a career in investment banking. if 2 students are chosen at random, what is the probability that at least one of them is interested in investment banking?
1/14 4/49 2/7 45/49 13/14 Easy way: None are interested in I banking. 5/7*21= 15 > 2115 = 6 6/21*5/20 > 2/7*1/4 > 2/28> 1/14 so simple 11/14 > 13/14 E



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Re: complementary events  probabilities [#permalink]
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06 Feb 2008, 09:04
Can one explain where I botched this....
I did Prob of one interested + Prob of both interested
(10/21)(11/20) + (10/21)(9/20) = 10/21 (Prob of interested)(Prob not interested) + (Prob interested)(Prob interested) = 10/21
Where did I screw this up?



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Re: complementary events  probabilities [#permalink]
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06 Feb 2008, 09:11
jimmyjamesdonkey wrote: Can one explain where I botched this....
I did Prob of one interested + Prob of both interested
(10/21)(11/20) + (10/21)(9/20) = 10/21 (Prob of interested)(Prob not interested) + (Prob interested)(Prob interested) = 10/21
Where did I screw this up? it's not 10 and 11 but 6 and 15



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Re: complementary events  probabilities [#permalink]
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06 Feb 2008, 09:14
Still not coming out right....
Now getting
(15/21)(6/20) + (15/21)(14/20) = 60/84. Anyone explain this?



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Re: complementary events  probabilities [#permalink]
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26 Jul 2013, 12:55
jimmyjamesdonkey wrote: Still not coming out right....
Now getting
(15/21)(6/20) + (15/21)(14/20) = 60/84. Anyone explain this? Can anyone explain why and how this approach is wrong? Thanks!



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Re: complementary events  probabilities [#permalink]
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26 Jul 2013, 13:39



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According to a recent student poll, 5/7 out of 21 members of [#permalink]
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23 Sep 2015, 11:14
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probability that first person is not interested in investment banking = 6/21 probability that second person is not interested in investment banking given that first also is not interested = 5/20 probability that at least one of them is interested in investment banking = 1  probability neither one of them interested in investment banking = 1  ((6/21) *(5/20)) = 13/14
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Re: According to a recent student poll, 5/7 out of 21 members of [#permalink]
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01 Mar 2018, 13:17
Hi All, When it comes to probability questions, there are 2 things that you can calculate: what you WANT and what you DON'T WANT. In probability, (the probability of what you WANT) + (the probability of what you DON'T WANT) = 1 In this question, we WANT AT LEAST 1 member chosen to be interested in investment banking; what we DON'T WANT is 0 members chosen to be interested in investment banking. The second option will be easier to calculate. Here's how: We're told that 5/7 of the 21 members are interested in investment banking: 15 interested in investment banking 6 NOT interested in investment banking We're asked to select 2 at random. Based on the above probability concepts…. 1  (probability that the 2 DON'T WANT investment banking) = the probability of AT LEAST 1 that does want investment banking The probability that the 1st DOESN'T WANT investment banking = 6/21 The probability that the 2nd DOESN'T WANT investment banking = 5/20 (6/21)(5/20) = 30/420 = 3/42 = 1/14 1  1/14 = 13/14 = the probability that AT LEAST 1 of the 2 chosen is interested in investment banking. Final Answer: GMAT assassins aren't born, they're made, Rich
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According to a recent student poll, 5/7 out of 21 members of [#permalink]
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12 May 2018, 04:06
OFFICIAL SOLUTION Solution: E “The probability of at least one” is very complicated to determine – you would need to count every time it happens and then divide it by the total number of possibilities. There is a much easier approach: calculating the inverse probability and then subtract it from one to get the actual probability.\((Remember: PA = 1 – P(Not A).)\) The logical inverse of “at least one” is “none at all.” For purposes of this problem, “none at all” means the probability of not picking one of the students interested in investment banking. \(\frac{5}{7}\) of the 21 total members of the finance club (or 15 members) are interested in investment banking. Thus, the chance of not picking an investment banker the first time around is \(\frac{6}{21}\) (since 6 are not interested.) Once you have picked a noninvestment banker out of the mix, there are five remaining noninvestment bankers out of 20 total left. Therefore, the chance of picking another noninvestment banker is \(\frac{5}{20}\). Since we need to pick twice and have both of our choices be noninvestment bankers, we must multiply the probabilities. The total inverse probability would be: \(P(inverse)=\frac{6}{21}∗\frac{5}{20}=\frac{2}{7}∗\frac{1}{4}=\frac{1}{14}\) Notice that this is an option (answer choice “A”). To finish off this problem, you need to transform the inverse probability into the actual, by subtracting the inverse probability from 1. Therefore, \(P(actual)=1−\frac{1}{14}=\frac{14}{14}−14=\frac{13}{14}\) The answer is “E”.
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