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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] New post 17 Oct 2017, 22:17
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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
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Re: From a group of 5 managers (Joon, Kendra, Lee, Marnie and No [#permalink] New post 18 Oct 2017, 05:11
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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)
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Re: From a group of 5 managers (Joon, Kendra, Lee, Marnie and No [#permalink] New post 04 Jan 2018, 08:38
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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.

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

Answer: A

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

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