It is currently 21 Jan 2018, 06:55

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

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

Author Message
TAGS:

### Hide Tags

VP
Status: Preparing for the GMAT
Joined: 02 Nov 2016
Posts: 1341

Kudos [?]: 1181 [1], given: 555

Location: Pakistan
GPA: 3.4
At a medical research lab, nine doctors are conducting multiple clinic [#permalink]

### Show Tags

17 Sep 2017, 10:34
1
KUDOS
3
This post was
BOOKMARKED
00:00

Difficulty:

95% (hard)

Question Stats:

35% (02:55) correct 65% (07:32) wrong based on 43 sessions

### HideShow timer Statistics

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

_________________

Official PS Practice Questions
Press +1 Kudos if this post is helpful

Kudos [?]: 1181 [1], given: 555

Intern
Joined: 28 Dec 2010
Posts: 49

Kudos [?]: 12 [3], given: 9

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

### Show Tags

23 Sep 2017, 22:58
3
KUDOS
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.

_______________
_________________

_________________________________________

Kudos [?]: 12 [3], given: 9

At a medical research lab, nine doctors are conducting multiple clinic   [#permalink] 23 Sep 2017, 22:58
Display posts from previous: Sort by