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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
2
3
00:00
A
B
C
D
E

Difficulty:

  95% (hard)

Question Stats:

43% (02:26) correct 57% (06:24) wrong based on 61 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

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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
4
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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Re: At a medical research lab, nine doctors are conducting multiple clinic [#permalink]

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New post 22 Apr 2018, 07:48
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SajjadAhmad wrote:
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


OA: E
As per Question stem , There are 3 Groups having 2 doctors each , 1 group having 3.
Total :4 groups , 9 Doctors. Lets us name these groups P,Q,R,S.
P: 2 Doctors
Q: 2 Doctors
R: 2 Doctors
S: 3 Doctors
Now we have to select 2 doctors such that these 2 are not working together.
Case 1:Numbers of ways of selecting 1 doctor from P,Selecting 1 doctor from Q = \(2*2 or C(2,1)*C(2,1)\)
Case 2:Numbers of ways of selecting 1 doctor from P,Selecting 1 doctor from R = \(2*2 or C(2,1)*C(2,1)\)
Case 3:Numbers of ways of selecting 1 doctor from Q,Selecting 1 doctor from R = \(2*2 or C(2,1)*C(2,1)\)

Case 4:Numbers of ways of selecting 1 doctor from P, Selecting 1 doctor from S = \(2*3 or C(2,1)*C(3,1)\)
Case 5:Numbers of ways of selecting 1 doctor from Q, Selecting 1 doctor from S = \(2*3 or C(2,1)*C(3,1)\)
Case 6:Numbers of ways of selecting 1 doctor from R, Selecting 1 doctor from S = \(2*3 or C(2,1)*C(3,1)\)

Total no ways of selecting 2 doctors such that these 2 are not working together. \(= 2*2+2*2+2*2+2*3+2*3+2*3 = 4+4+4+6+6+6\)
\(= 12+18 =30\)
Total no ways of selecting 2 doctors out of 9 , without any precondition = C(9,2) = \(\frac{9*8}{2}=36\)

Probability = \(\frac{30}{36}\)=\(\frac{10}{12}\)
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Re: At a medical research lab, nine doctors are conducting multiple clinic   [#permalink] 22 Apr 2018, 07:48
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At a medical research lab, nine doctors are conducting multiple clinic

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