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At a medical research lab, nine doctors are conducting multiple clinic

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At a medical research lab, nine doctors are conducting multiple clinic [#permalink]

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New post 17 Sep 2017, 11:34
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A
B
C
D
E

Difficulty:

  95% (hard)

Question Stats:

35% (02:41) correct 65% (01:52) wrong based on 37 sessions

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At a medical research lab, nine doctors are conducting multiple clinical trials. Six of the doctors are working on a clinical trial with exactly one other doctor and three doctors are working on a clinical trial with exactly two other doctors. If two doctors are selected at random from the lab, what is the probability that those two doctors are NOT working together on a clinical trial?

A 1/12
B 2/12
C 5/12
D 7/12
E 10/12
[Reveal] Spoiler: OA

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At a medical research lab, nine doctors are conducting multiple clinic [#permalink]

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New post 23 Sep 2017, 23:58
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Lets first understand the question step, and call doctors with names : A,B,C... I
Here are the groups:
AB
CD
EF
GHI

Now, to find the probability that two doctors are NOT working together on a clinical trial, we can find probability of 2 doctors working together and subtract it from 1.

Probability (of 2 docs working together) = \(\frac{(2 docs from first 3 groups + any 2 docs from fourth group)}{select any 2 docs from 9}\)

Probability (of 2 docs working together) = \(\frac{3 + 3C2}{9C2}\)

Probability (of 2 docs working together) = \(\frac{6}{36}\)

Probability (of 2 docs working together) = \(\frac{1}{6}\)

Now Probability (not selecting 2 docs) = 1 - Probability (of 2 docs working together)
= 1 - \(\frac{1}{6}\)
= \(\frac{5}{6}\)

E.

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At a medical research lab, nine doctors are conducting multiple clinic   [#permalink] 23 Sep 2017, 23:58
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