Problem Solving

razaul karim 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
Ans:
This means that P(M and N both selected) = [spoiler](2/5) x (1/4) = 1/10[/spoiler]

Another question::
Joshua and Jose work at an auto repair center with 4 other workers. For a survey on health care insurance, 2 of the 6 workers will be randomly chosen to be interviewed. What is the probability that Joshua and Jose will both be chosen?

a)1/15
b)1/12
c)1/9
d)1/6
e)1/3

Ans:
[spoiler](1/6)*(1/5)+(1/6)*(1/5)=2/30=1/15[/spoiler]
Here why do we consider Both the options (Jose*Joshuna) and (Joshuna*jose)
But in the first question we considerded probability just 2/5x1/4 and why not (2/5x1/4)(1/4x2/5) like we did in the second problem.Please explain the difference..
Please explain

sanaa.rizwan wrote:I am not getting the solution. why is the probability of the first occurrence/selection 2/5 and not 1/5

When 2 events occur one after the other, we are suppose to multiply their probabilities.

The probability that Marnie is selected is 1/5
The probability that Noomi is selected is 1/4 (since only 4 employees are left)

so the probability that both Marnie and Noomi are selected is 1/5 * 1/4 = 1/20

This is the method used in one of the MGMAT practise test explanations.

Where am I going wrong?

razaul karim 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