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# From a group of 5 managers (Joon, Kendra, Lee, Marnie and No

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From a group of 5 managers (Joon, Kendra, Lee, Marnie and No [#permalink]  17 Oct 2017, 22:17
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Question Stats:

55% (00:45) correct 44% (01:14) wrong based on 9 sessions
From a group of 5 managers (Joon, Kendra, Lee, Marnie and Noomi), 2 people are randomly selected to attend a conference in Las Vegas. What is the probability that Marnie and Noomi are both selected?

(A) 0.1

(B) 0.2

(C) 0.25

(D) 0.4

(E) 0.6

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[Reveal] Spoiler: OA
Director
Joined: 03 Sep 2017
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Kudos [?]: 327 [1] , given: 66

Re: From a group of 5 managers (Joon, Kendra, Lee, Marnie and No [#permalink]  18 Oct 2017, 05:11
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KUDOS
There are 10 ways to select 2 people out of 5 $$\frac{5!}{2!3!}$$.

Then, there is just one way to select Marni and Noomi.

Thus, the probability of choosing them is 1/10 = 0.1 (A)
Target Test Prep Representative
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Re: From a group of 5 managers (Joon, Kendra, Lee, Marnie and No [#permalink]  04 Jan 2018, 08:38
Expert's post
Bunuel wrote:
From a group of 5 managers (Joon, Kendra, Lee, Marnie and Noomi), 2 people are randomly selected to attend a conference in Las Vegas. What is the probability that Marnie and Noomi are both selected?

(A) 0.1

(B) 0.2

(C) 0.25

(D) 0.4

(E) 0.6

There are a total of 5C2 = 5!/(3! x 2!) = (5 x 4)/2 = 10 ways of choosing two people from a group of five people. Since the selection of Marnie and Noomi corresponds to one of these choices, the probability of this selection is 1/10 = 0.1.

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Re: From a group of 5 managers (Joon, Kendra, Lee, Marnie and No   [#permalink] 04 Jan 2018, 08:38
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