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A class contains five juniors and five seniors. If one member of the

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Bunuel
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A class contains five juniors and five seniors. If one member of the [#permalink] New post   28 Aug 2015, 02:57
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A class contains five juniors and five seniors. If one member of the class is assigned at random to present a paper on a certain subject, and another member of the class is randomly assigned to assist him, what is the probability that both will be juniors?

A. 1/10
B. 1/5
C. 2/9
D. 2/5
E. 1/2


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C
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noTh1ng
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Re: A class contains five juniors and five seniors. If one member of the [#permalink] New post   28 Aug 2015, 05:29
I might be wrong:
but what is the matter with

P(junior) * P(Junior) = 5/10 * 4/9 = 2/9 ?
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Re: A class contains five juniors and five seniors. If one member of the [#permalink] New post   Updated on: 28 Aug 2015, 12:28
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This is an interesting probability problem.Please correct me if I am wrong

The probability of first assigned student as Junior =Number of Juniors /Total Students =5/10=1/2




The probability of Second assigned student as Junior = Number of remaining junior students/Total remaining Students=4/9


So the probablity of both students assigned as Junior=(1/2) * (4/9)=2/9



So the Correct Answer C

Originally posted by AbdurRakib on 28 Aug 2015, 11:16.
Last edited by AbdurRakib on 28 Aug 2015, 12:28, edited 1 time in total.
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Re: A class contains five juniors and five seniors. If one member of the [#permalink] New post   28 Aug 2015, 11:55
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noTh1ng wrote:
I might be wrong:
but what is the matter with

P(junior) * P(Junior) = 5/10 * 4/9 = 2/9 ?


I think you're right. The other way to do it (but the long way) is to figure out the probability that it is not two Juniors.

2 seniors = P(Senior) * P(Senior) = 2/9

1 Senior and 1 Junior = (1/2) *(5/9)*2 = 5/9

Probability that it is not two Juniors is 5/9+2/9 = 7/9 so the probability that it is two juniors is 1- (7/9) = 2/9.

Again unnecessarily long but it does check out
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Re: A class contains five juniors and five seniors. If one member of the [#permalink] New post   01 Sep 2015, 19:31
Wofford09 wrote:
noTh1ng wrote:
I might be wrong:
but what is the matter with

P(junior) * P(Junior) = 5/10 * 4/9 = 2/9 ?


I think you're right. The other way to do it (but the long way) is to figure out the probability that it is not two Juniors.

2 seniors = P(Senior) * P(Senior) = 2/9

1 Senior and 1 Junior = (1/2) *(5/9)*2 = 5/9

Probability that it is not two Juniors is 5/9+2/9 = 7/9 so the probability that it is two juniors is 1- (7/9) = 2/9.

Again unnecessarily long but it does check out

Hi Wofford09,

I think your solution is correct. However, and as you stated, I don´t think this is the fastest approach.
This is my point: what you are doing follows the reasoning of "Desired outcome = 1 - Undesired outcome". This approach is a shortcut when the undesired outcome is easier to calculate than the desired. In this case, it is easier to get to the desired outcome with the simple calculation (1/2)(4/9) = 2/9.

Nonetheless, it´s great to know both methods because they will come in handy at 700+ problems. So thanks for pointing it out!

Cheers,

=)
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Re: A class contains five juniors and five seniors. If one member of the [#permalink] New post   02 Sep 2015, 00:16
Apparently the solution is 2/5?

Bunuel can you help?
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Re: A class contains five juniors and five seniors. If one member of the [#permalink] New post   02 Sep 2015, 05:50
I would like to see an explanation as well. I just googled to see if anyone else had an issue with this question and Kaplan shows the solution as 2/9 in their ACT guide book.


Thanks
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Re: A class contains five juniors and five seniors. If one member of the [#permalink] New post   02 Sep 2015, 05:56
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Wofford09 wrote:
I would like to see an explanation as well. I just googled to see if anyone else had an issue with this question and Kaplan shows the solution as 2/9 in their ACT guide book.


Thanks


The OA is 2/9. Edited.
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Re: A class contains five juniors and five seniors. If one member of the [#permalink] New post   02 Sep 2015, 06:09
Bunuel wrote:
A class contains five juniors and five seniors. If one member of the class is assigned at random to present a paper on a certain subject, and another member of the class is randomly assigned to assist him, what is the probability that both will be juniors?

A. 1/10
B. 1/5
C. 2/9
D. 2/5
E. 1/2


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number of ways of selecting 2 members our of 10 members - 10C2
number of ways of selecting 2 juniors out of 5 - 5C2
probability that both members are juniors is

\(\frac{5C2}{10C2}\)

or \(\frac{5*4}{10*9}\)

or \(\frac{2}{9}\)

Answer:- C
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Re: A class contains five juniors and five seniors. If one member of the [#permalink] New post   13 Nov 2020, 01:11
I solved it as under:-

Total ways of picking 2 students out of 10 = 10*9 = 90
Total ways of picking 2 juniors only = 1C5 * 1C4 = 20

P (2 juniors only) = 20/90 = 2/9

I'm very weak in P&C, so took the long route. But yeah, one can also just do 2C5/2C10 =10/45 = 2/9
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Re: A class contains five juniors and five seniors. If one member of the [#permalink]
13 Nov 2020, 01:11
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