A person is to be selected at random from the group T of people...

How can we assume that club D and club E are the only clubs that T can be a part of ?

Is this indicative of a typical GMAT question? I've not come across a question in the OG where I have had to make such an assumption. (That D and E are the only clubs in T) Usually, it is mentioned in the question, which would make the answer E.

Hey thanks for raising your query. The source of this question is Kaplan. I wouldn't say the wording is incorrect. Let me copy paste statement 1:-

1. The probability that a person selected at random from group T is not a member of club D and is not a member of club E is 1/4.

What do you infer from statement one?

Does your concern hold true any longer?

I have a similar doubt. It is not considered the possibility that a person is member of club D and E at the same time, which would change the overall probability of being part of club D.
I think it should be explicitly stated that a person can't be part of club D and E at the same time, otherwise answer must be E.

Hello

Lets take A= event that person is ONLY a member of club D
B = event that person is ONLY a member of club E
C= event that person is a member of BOTH clubs D/E
D= event that person is neither a member of club D nor a member of club E

We have to find the probability that person is a member of club E, which would include both events B&C. So the combined probability of B&C: P(B) + P(C).
And we know that P(A)+P(B)+P(C)+P(D) = 1 (total probability)

Statement 1 gives us probability of event D, P(D).
Statement 2 gives us probability of event A, P(A).
So combining, we can easily get P(B)+P(C).

I have a similar doubt. It is not considered the possibility that a person is member of club D and E at the same time, which would change the overall probability of being part of club D.
I think it should be explicitly stated that a person can't be part of club D and E at the same time, otherwise answer must be E.[/quote]


Hello

Lets take A= event that person is ONLY a member of club D
B = event that person is ONLY a member of club E
C= event that person is a member of BOTH clubs D/E
D= event that person is neither a member of club D nor a member of club E

We have to find the probability that person is a member of club E, which would include both events B&C. So the combined probability of B&C: P(B) + P(C).
And we know that P(A)+P(B)+P(C)+P(D) = 1 (total probability)

Statement 1 gives us probability of event D, P(D).
Statement 2 gives us probability of event A, P(A).
So combining, we can easily get P(B)+P(C).[/quote]

Okay, now it's clear. Thank you so much, I was looking at the incorrect data.

Either statement alone won't be sufficient, as we'll have two variables and single equation.
Taking both statements together:

Statement 1: p[not D].p[not E] = 1/4
Statement 2:
p[D].p[not E] = 5/12
or, {1-p[not D]}.p[not E] = 5/12
or, p[not E] - p[not D].p[not E] = 5/12
or, p[not E] - 1/4 = 5/12
or, p[not E] = 8/12 = 2/3

Therefore, p[E] = 1 - p[not E] = 1 - 2/3 = 1/3.

Answer: C

I feel both statements combined will still not be sufficient. Because there is no way by which we know that there are only two clubs D and E and not any other.

Cheers,
PV66

That point gets covered in statement I. There is 1/4 chance that the person is from all other clubs ( less D and E)

Thanks Chetan, guess I didn't read statement 1 that carefully. It actually makes sense once both statements are combined.

Regards,
PV66

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