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Bunuel
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From a group of 5 managers (Joon, Kendra, Lee, Marnie and Noomi), 2 13 Sep 2017, 23:20
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A
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35% (medium)

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66% (01:15) correct 34% (01:20) wrong based on 438 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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A
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From a group of 5 managers (Joon, Kendra, Lee, Marnie and Noomi), 2 13 Sep 2017, 23:32
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Bunuel
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
Show more

The total ways of choosing 2 people from a group of 5 managers are \(5c2(10)\)

If we need to find the probability that Marnie and Noomi are chosen, that's just 1 of the 10 combinations possible.

Hence, the probability that both Marnie and Noomi are chosen to attend the conference in Las Vegas is \(\frac{1}{10}\) = 0.1(Option A)
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From a group of 5 managers (Joon, Kendra, Lee, Marnie and Noomi), 2 14 Sep 2017, 02:29
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pushpitkc
Bunuel
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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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?
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From a group of 5 managers (Joon, Kendra, Lee, Marnie and Noomi), 2 18 Sep 2017, 05:47
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Bunuel
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
Show more

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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From a group of 5 managers (Joon, Kendra, Lee, Marnie and Noomi), 2 18 Sep 2017, 16:06
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Bunuel
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
Show more

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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From a group of 5 managers (Joon, Kendra, Lee, Marnie and Noomi), 2 14 Jun 2023, 03:29
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