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probability

This topic has 3 expert replies and 2 member replies

probability

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

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:
(1/6)*(1/5)+(1/6)*(1/5)=2/30=1/15
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

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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) = (2/5) x (1/4) = 1/10

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:
(1/6)*(1/5)+(1/6)*(1/5)=2/30=1/15
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
The second question can be solved using the exact same approach used in the first question.
That is, P(Joshua and Jose both chosen) = (2/6) x (1/5) = 1/15

The solution you posted still has the correct answer (1/15). The only difference is that it considers two acceptable outcomes:
1) Joshua chosen 1st and Jose chosen 2nd
2) Jose chosen 1st and Joshua chosen 2nd

Cheers,
Brent

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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?

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GMAT/MBA Expert

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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?
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
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

Both solutions yield the same answer.

Your approach, seems to consider one possible outcome (M selected 1st AND N selected 2nd) and doesn't consider the other one (N selected 1st AND M selected 2nd)

Cheers,
Brent

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Now I get it why is is 2/5 and not 1/5.

Thanks a lot

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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
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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