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From a group of 5 managers (Joon, Kendra, Lee, Marnie and No
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17 Oct 2017, 22:17
17 Oct 2017, 22:17
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Question Stats:
67% (00:55) correct
32% (01:16) wrong based on 28 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
Kudos for correct solution.
(A) 0.1
(B) 0.2
(C) 0.25
(D) 0.4
(E) 0.6
Kudos for correct solution.
Re: From a group of 5 managers (Joon, Kendra, Lee, Marnie and No
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18 Oct 2017, 05:11
18 Oct 2017, 05:11
1
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)
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
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04 Jan 2018, 08:38
04 Jan 2018, 08:38
1
Expert Reply
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
(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
