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

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Math Expert
Joined: 02 Sep 2009
Posts: 46284
From a group of 5 managers (Joon, Kendra, Lee, Marnie and Noomi), 2 [#permalink]

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13 Sep 2017, 23:20
00:00

Difficulty:

45% (medium)

Question Stats:

61% (00:49) correct 39% (00:41) wrong based on 81 sessions

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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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Posts: 2834
Location: India
GPA: 3.12
Re: From a group of 5 managers (Joon, Kendra, Lee, Marnie and Noomi), 2 [#permalink]

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13 Sep 2017, 23:32
1
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

The total ways of choosing 2 people from a group of 5 managers is $$5c2(10)$$
If we need to find the probability that Marnie and Noomi are chosen, thats is just 1 of the 10 combinations possible.

Hence, the probability that both Marnie and Noomi are chosen from the group
of managers to attend the conference in Las Vegas is $$\frac{1}{10}$$ = 0.1(Option A)
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Re: From a group of 5 managers (Joon, Kendra, Lee, Marnie and Noomi), 2 [#permalink]

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14 Sep 2017, 02:29
pushpitkc wrote:
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

The total ways of choosing 2 people from a group of 5 managers is $$5c2(10)$$
If we need to find the probability that Marnie and Noomi are chosen, thats is just 1 of the 10 combinations possible.

Hence, the probability that both Marnie and Noomi are chosen from the group
of managers to attend the conference in Las Vegas is $$\frac{1}{10}$$ = 0.1(Option A)

Hi,

Can you please let me know what is wrong in this approach-

P (each getting picked) = 1/5
Now as per question stem, since we want both Marnie and Noomi to be selected , then -

P (Marnie or Noomi) * P (the other one) = 1/5 * 1 = 0.2?
Target Test Prep Representative
Affiliations: Target Test Prep
Joined: 04 Mar 2011
Posts: 2570
Re: From a group of 5 managers (Joon, Kendra, Lee, Marnie and Noomi), 2 [#permalink]

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18 Sep 2017, 05:47
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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18 Sep 2017, 16:06
Top Contributor
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] 18 Sep 2017, 16:06
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