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

Question

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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Official Answer

E

A 1/12
B 2/12
C 5/12
D 7/12
E 10/12

Are You Up For the Challenge: 700 Level Questions

false9 · Answer

IMO E.

Question says there are total 9 doctors. Out of these 9, 6 are working on clinical trial with exactly one another doctor and 3 are working on a clinical trial with exactly two other doctors. So we can divide these 9 doctors into 2 groups. One group of 6 and other group of 3.
First group consisting 6 (P,Q,R,S,T,U) will have doctors working on clinical trial with one another doctor (Say P with Q, R with S, and T with U). Second group consisting 3 (Say, V,W,X) will have exactly two other doctors as partners.

Now coming to the question. Better to approach by complement method i.e. via calculating the probability that 2 people selected out of the 9 are working together on a clinical trial.
In that case 1) we may select 2 out of second group (Out of V,W,X) in 3 ways.
OR 2) We can select 2 out of First group (Out of P,Q,R,S,T,U) in 3 ways. (P-Q, R-S, T-U)
So total 6 case we have in which we can select 2 people, both working on same clinical trial.

Number of ways in which 2 people can be selected out of 9 = 9c2 = 36.

Probability that selected 2 work on same clinical trial = 6/36 = 1/6.
Probability that selected 2 do not work on same clinical trial = 1 - (1/6) = 5/6. (E)

Feedbacks would be appreciated.

aryakanya10110 · Answer

your ans is 5/6 not in opt.

y 6/36 -?
y not 3/36

Regor60 · Answer

Two approaches.

First: direct probability

There are 6 members in the first group and 3 in the second.

The probability of a doctor being selected from the first group is

6/9=2/3

Since there are 8 remaining doctors to choose from, but the first doctor's partner is disqualified, the probability of picking a non partner is

7/8

So the probability of two doctors being picked that aren't partners where 1 doctor is picked from the first group is
 
 2/3*7/8 = 14/24

There is a complementary chance of 1/3 of picking the doctor from the second group of 3

With 8 doctors again remaining, but now 2 doctors are disqualified, the probability of picking a non partner is

6/8

So the probability of 2 doctors being picked that aren't partners where 1 doctor is picked from the second group is

1/3*6/8 = 6/24

The overall probability is then

14/24 + 6/24 = 20/24 = 10/12

Using combinations

The number of ways to pick a doctor from the first group

6

The number of ways to then pick a non partner

7

Total ways 6*7/2 = 21. Divide by 2 because order doesn't matter.

Number of ways to pick a doctor from the second group

3

Number of ways to pick a non partner

6

Total ways 3*6/2 = 9, again dividing by 2 because order doesn't matter

So total ways is

21+9 = 30

To compute probability, need total ways 2 doctors can be chosen from 9

9!/2!7! = 36

Probability: 30/36 = 10/12

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thkkrpratik · Answer

In this part, won't it be more than 3 ways, for example - PQ, PR, PS, PT, PU, QR, QS, QT, QU,...?

Am I missing something here?

Bunuel , when will the OA be published?

GMATGuruNY · Answer

Let's say that the 9 doctors are ABCDEFGHI.

Six of the doctors are working on a clinical trial with exactly one other doctor. 
Let's say that A and B are working together, that C and D are working together, and that E and F are working together.

Three doctors are working on a clinical trial with exactly two other doctors .
Let's say that G, H and I are all working with one another.

Total number of doctors pairs that are working together = 6 (AB, CD, EF, GH, GI, and HI)
Total number of pairs that can be formed from 9 doctors = 9C2 =  = \frac{9*8}{2*1} = 36 
P(random pair is working together) = 6/36 = 1/6
P(random pair is NOT working together = 1 - 1/6 = 5/6 = 10/12

E

Regor60 · Answer

You're picking two that work together, not any two, which would be a combination.

There are only 3 sets of 2 that work together.

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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

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