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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