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

AAPL wrote:Manhattan Prep

From a group of 5 managers (Joo, 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

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

AAPL wrote:Manhattan Prep

From a group of 5 managers (Joo, 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

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