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From a group of 5 managers (Joon, Kendra, Lee, Marnie and

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From a group of 5 managers (Joon, Kendra, Lee, Marnie and

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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

OA A

Source: Manhattan Prep

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BTGmoderatorDC 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

OA A

Source: Manhattan Prep
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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GMAT/MBA Expert

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BTGmoderatorDC 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

OA A

Source: Manhattan Prep
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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Post
BTGmoderatorDC 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
Approach 1;
P = (good pair)/(all possible pairs)
Good pair = Marnie and Noomi = 1 option.
All possible pairs = the number of ways to select 2 managers from 5 options = 5C2 = (5*4)/(2*1) = 10 options.
Thus:
P = 1/10 = 0.1

Approach 2:
P(M or N is selected first) = 2/5. (Of the 5 managers, 2 are M or N.)
P(M or N is selected second) = 1/4. (Of the 4 remaining managers, 1 will be M or N, since either M or N was selected first.)
To combine these probabilities, we multiply:
2/5 * 1/4 = 1/10 = 0.1

The correct answer is A.

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