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Jack and Jill work at a hospital with 4 other workers. For a

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Jack and Jill work at a hospital with 4 other workers. For a  [#permalink]

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New post Updated on: 14 Jun 2013, 11:29
1
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C
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Jack and Jill work at a hospital with 4 other workers. For an internal review, 2 of the 6 workers will be randomly chosen to be interviewed. What is the probability that Jack and Jill will both be chosen?

(A) 1/3
(B) 1/4
(C) 1/15
(D) 3/8
(E) 2/3

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Originally posted by atalpanditgmat on 14 Jun 2013, 11:19.
Last edited by Bunuel on 14 Jun 2013, 11:29, edited 2 times in total.
OA added.
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Re: Jack and Jill work at a hospital with 4 other workers. For a  [#permalink]

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New post 14 Jun 2013, 15:50
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2
Total number of people = 6

Probability of selecting Jack first and then Jill is \(\frac{1}{6}\)\(*\frac{1}{5}\)\(=\frac{1}{30}\)
Probability of selecting Jill first and then Jack is \(\frac{1}{6}\)\(*\frac{1}{5}\)\(=\frac{1}{30}\)

Therefore probability of selecting Jack and Jill for the review is \(\frac{1}{30}\)\(+\frac{1}{30}\)\(=\frac{1}{15}\)

Answer is C
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Re: Jack and Jill work at a hospital with 4 other workers. For a  [#permalink]

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New post 14 Jun 2013, 11:30
2
atalpanditgmat wrote:
Jack and Jill work at a hospital with 4 other workers. For an internal review, 2 of the 6 workers will be randomly chosen to be interviewed. What is the probability that Jack and Jill will both be chosen?

(A) 1/3
(B) 1/4
(C) 1/15
(D) 3/8
(E) 2/3


1/6C2=1/15.

Answer: C.
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Re: Jack and Jill work at a hospital with 4 other workers. For a  [#permalink]

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New post 25 Oct 2015, 13:35
Bunuel wrote:
atalpanditgmat wrote:
Jack and Jill work at a hospital with 4 other workers. For an internal review, 2 of the 6 workers will be randomly chosen to be interviewed. What is the probability that Jack and Jill will both be chosen?

(A) 1/3
(B) 1/4
(C) 1/15
(D) 3/8
(E) 2/3


1/6C2=1/15.

Answer: C.


I just did chance of 1 being selected first time is 1/6 + chance of 1 being selected second time 1/5, turns into 2/30 and reduces to 1/15. Is that the same as 1/6C2=1/15 (my math fundamentals are still lacking a bit)?
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Re: Jack and Jill work at a hospital with 4 other workers. For a  [#permalink]

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New post 25 Oct 2015, 15:07
2
redfield wrote:
Bunuel wrote:
atalpanditgmat wrote:
Jack and Jill work at a hospital with 4 other workers. For an internal review, 2 of the 6 workers will be randomly chosen to be interviewed. What is the probability that Jack and Jill will both be chosen?

(A) 1/3
(B) 1/4
(C) 1/15
(D) 3/8
(E) 2/3


1/6C2=1/15.

Answer: C.


I just did chance of 1 being selected first time is 1/6 + chance of 1 being selected second time 1/5, turns into 2/30 and reduces to 1/15. Is that the same as 1/6C2=1/15 (my math fundamentals are still lacking a bit)?



Yes. nCr = n!/(r!*(n-r)!), thus 6C2 = 6!/(2!*4!) = 15

Thus 1/6C2 = 1/15

The correct way to approach is a follows:

Selecting 1 out of the 2 = 2/6 and then the remaining 1 person out of Jack and Jill = 1/5 , giving you the final probability = 2/6*1/5 = 1/15

Hope this helps.
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Jack and Jill work at a hospital with 4 other workers. For an internal  [#permalink]

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New post 02 Jun 2016, 04:33
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Re: Jack and Jill work at a hospital with 4 other workers. For an internal  [#permalink]

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New post 02 Jun 2016, 06:41
2
Bunuel wrote:
Jack and Jill work at a hospital with 4 other workers. For an internal review, 2 of the 6 workers will be randomly chosen to be interviewed. What is the probability that Jack and Jill will both be chosen?

(A) 1/3
(B) 1/4
(C) 1/15
(D) 3/8
(E) 2/3


Total number of ways to choose 2 out of 6 workers= 6!/2!4!= 15

Number of ways to choose both jack and jill= 1

probability= 1/15

C should be the answer
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Re: Jack and Jill work at a hospital with 4 other workers. For an internal  [#permalink]

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New post 02 Jun 2016, 11:16
Probability that Jack and Jill will both be chosen out of 6 workers
= (2/6)*(1/5)
= 1/15
Answer C
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Re: Jack and Jill work at a hospital with 4 other workers. For a  [#permalink]

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