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# From a group of 5 managers (Joon, Kendra, Lee, Marnie and Noomi), 2

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Re: From a group of 5 managers (Joon, Kendra, Lee, Marnie and Noomi), 2 [#permalink]
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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.

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Re: From a group of 5 managers (Joon, Kendra, Lee, Marnie and Noomi), 2 [#permalink]
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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

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

There are two ways to approach this.

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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Re: From a group of 5 managers (Joon, Kendra, Lee, Marnie and Noomi), 2 [#permalink]
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Re: From a group of 5 managers (Joon, Kendra, Lee, Marnie and Noomi), 2 [#permalink]
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