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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:

60% (00:46) correct 40% (01:23) wrong based on 15 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

Kudos for correct solution.
[Reveal] Spoiler: OA
Director
Joined: 03 Sep 2017
Posts: 520
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Kudos [?]: 356 [1] , given: 66

Re: From a group of 5 managers (Joon, Kendra, Lee, Marnie and No [#permalink]  18 Oct 2017, 05:11
1
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
Affiliations: Target Test Prep
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Kudos [?]: 140 [1] , given: 0

Re: From a group of 5 managers (Joon, Kendra, Lee, Marnie and No [#permalink]  04 Jan 2018, 08:38
1
KUDOS
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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Kudos [?]: 1803 [1] , given: 9

Re: From a group of 5 managers (Joon, Kendra, Lee, Marnie and No [#permalink]  21 Apr 2019, 07:17
1
KUDOS
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

Kudos for correct solution.

Our goal is to find P(M and N both selected)

Method #1:
P(M and N both selected) = P(one of them is selected 1st AND the other selected 2nd)
= P(one of them is selected 1st) x P(the other selected 2nd)
= (2/5)(1/4)
= 1/10
= 0.1
Aside: P(one of them is selected 1st) = 2/5 because I'm allowing for either Marnie or Noomi to be selected first.

Method #2:
P(M and N both selected) = P(M selected 1st AND N selected 2nd OR N selected 1st AND M selected 2nd)
= P(M selected 1st AND N selected 2nd) + P(N selected 1st AND M selected 2nd)
= (1/5)(1/4) + (1/5)(1/4)
= 1/20 + 1/20
= 1/10
= 0.1

Cheers,
Brent
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Brent Hanneson – Creator of greenlighttestprep.com

Re: From a group of 5 managers (Joon, Kendra, Lee, Marnie and No   [#permalink] 21 Apr 2019, 07:17
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