When 100 people who have not used cocaine are tested for cocaine use

Question

When 100 people who have not used cocaine are tested for cocaine use, on average only 5 will test positive. By contrast, of every 100 people who have used cocaine 99 will test positive. Thus, when a randomly chosen group of people is tested for cocaine use, the vast majority of those who test positive will be people who have used cocaine.

A reasoning error in the argument is that the argument

(A) attempts to infer a value judgment from purely factual premises
(B) attributes to every member of the population the properties of the average member of the population
(C) fails to take into account what proportion of the population have used cocaine
(D) ignores the fact that some cocaine users do not test positive
(E) advocates testing people for cocaine use when there is no reason to suspect that they have used cocaine

IanStewart · Accepted Answer

I've seen a few similar questions about 'false positives' and 'false negatives'. You can imagine the following situation:

100 people in Country X use cocaine
1,000,000 people in Country X do not use cocaine

Test all of these people, and 99 of the 100 cocaine users will test positive, while 50,000 of the non-users will test positive. You have 50,099 people in total who test positive, but of those, only 99/50,099 ~ 0.2 % are actually cocaine users.

Without knowing something about cocaine use in the population as a whole, we can't say very much about the people who will test positive. C is certainly the answer.

hanumayamma · Answer

Let us consider following give samples:
Sample A: Not Used Cocaine Users: Number of people reported positive: 5% (on average)
Sample B: Cocaine User: Number of people reported positive: 99%
As per argument, if you chosen randomly, the vast majority of those who test positive will be people who have used cocaine. Definitely the author would have taken Cocaine when he concluded this (Sorry for being rude - :roll:

). Nevertheless, author faultily reasoning all the samples will be from “Sample B”. That is, he didn’t take complete count of the two groups.

). Nevertheless, author faultily reasoning all the samples will be from “Sample B”. That is, he didn’t take complete count of the two groups. (A) attempts to infer a value judgment from purely factual premises ( If so, author reasoning would have been a correct one – eliminate it ) (B) attributes to every member of the population the properties of the average member of the population ( Enticing! Yes. This could be a potential winner (out of 200, 104 are Cocaine Users. ) However, the members were chosen randomly – Oops – Eliminate it ). (C) fails to take into account what proportion of the population have used cocaine( This is a good candidate too. But is this valid under random selection? Consider, the probability of picking a Cocaine User is 99 / 100 and picking non-Cocaine User is as Cocaine User = 5/100.So, no-way that we end up have same vast majority of Cocaine Users. Let twist the argument by saying 99% of non-Cocaine users will be tested positive. Then both probabilities will match. In short, if we have proper proportions, then we can accurately find the numbers that tested positive. – Hold it ) (D) ignores the fact that some cocaine users do not test positive ( may be, but does not provide vast majority – eliminate it ) (E) advocates testing people for cocaine use when there is no reason to suspect that they have used cocaine ( correct – but won’t address logical error – eliminate it ) Answer: C

chicagocubsrule · Answer

With Flaw questions always try to identify the flaw before moving to the answer choices. Here we are told because an average # of a particular group tests positive that the "vast majority of those who test positive will be people who have used cocaine". This is similar to saying since the average GPA in a 3rd year engineer students is a 3.0, the "vast majority" of engineers' GPA is a 3.0. What if 1st year engineer students average a 2.0? or a 4.0? You cannot make a conclusion about the average GPA of all engineer students based on a subset of the class.
This is nicely stated in B.

Technext · Answer

When 100 people who have not used cocaine are tested for cocaine use, on average only 5 will test positive. By contrast, of every 100 people who have used cocaine 99 will test positive. Thus, when a randomly chosen group of people is tested for cocaine use, the vast majority of those who test positive will be people who have used cocaine.

A reasoning error in the argument is that the argument

Explanation :
-----------------------
( A ) attempts to infer a value judgment from purely factual premises ---> In that case, it would have strengthen the argument.

( B ) attributes to every member of the population the properties of the average member of the population ---> At the most, this option might strengthen the argument. If everyone shares the same property, the conclusion will strengthen. So, discard it.

( C ) fails to take into account what proportion of the population have used cocaine 
---> This looks fine.

The passage discusses about two groups: 
 Group 1 . One in which, on an average, only 5 test positive 
 Group 2 . Other in which 99 test positive.

Conclusion makes a reasoning error in assuming that even if a  randomly chosen group  is tested, majority of them will be the ones who have used cocaine i.e., they will belong to group 2. This may not necessarily be true.

What if the  majority  of the randomly chosen group comprises people  belonging to group 1  ? Though they will still test positive, but they will belong to the group that doesn’t use cocaine. In that case, the argument will become weak.

( D ) ignores the fact that some cocaine users do not test positive ---> Irrelevant

( E ) advocates testing people for cocaine use when there is no reason to suspect that they have used cocaine. ---> Irrelevant.
---------------------

My choice is  C .

Hope that helps.

Regards,
Technext

Paris75 · Answer

Hi,

For me it is clearly C.

Why? Well because if 99.99999% Of 1000000 of a population never took cocaine but only 100 has. Than (1000000*0.05 = 50000 who did not take drug and you have on the other side 99 who take drug)

Than with this demonstration, the argument is totaly flawed.

You need to take into account the proportion or the part of the population actually taking drugs.

Answer C .

SVaidyaraman · Answer

Negate the conclusion:

It is: The vast majority of those who test positive out of a random group of people would  not  have used cocaine.

Since this a find the flaw question we have to select a choice which can weaken the conclusion .

So let us consider this as a weaken question first and so see which choice offers the best explanation for the negated conclusion.

Choice C because a particular case of it which is , a small proportion of the population take cocaine, offers the best explanation for the negated conclusion.

This is because assume 10000 people are tested . A small proportion of that population be 1% i.e, 100 people take cocaine. 99 of them would test positive. Of the remaining 9900 people 5%, i.e., 445 would test positive. So we see the majority of those who test positive are those who had not used cocaine.
i.e, if

Fdambro294 · Answer

Out of 100 people who don’t do cocaine: 5 test positive on average.

We can say 5% of (ppl who do NOT do cocaine test positive)

Out of 100 people who DO cocaine: 99 test positive on average

We can say that 99% of (ppl who DO cocaine will test positive)

Conclusion: the vast majority of those who TEST POSITIVE (from a randomly selected group) will be those who DO cocaine

This is similar to a weighted average set up in the quant section (something that the GMAT apparently loves to play around with).

Out of the entire population from which we are picking these random people, we do not know the “relative weighting” of the people who DO cocaine vs. the people who do NOT do cocaine.

For example, say we have 100 people in a hypothetical world.

99 do NOT do cocaine. Only 1 person does cocaine.

5% (99) = test positive but did NOT do cocaine

99% (1) = test positive and DID do cocaine

5% of 99 ~ 5 people tested positive who did NOT do cocaine

99% of 1 ~ 1 person who tested positive and actually DID the cocaine

Out of these 6 people who tested positive, the vast majority is those people who did NOT do cocaine (5 out of 6)

Because we do not know what proportion of the population actually does cocaine, we can’t make the conclusion that OUT OF those who TEST POSITIVE, most will have actually done cocaine

C

Posted from my mobile device

sachi-in · Answer

Prethinking will help in reducing time spent in over analyzing all the options.

Prethinking: 5 of 100 or 50 of 1000 will be tested false positive
 99 of 200 or 10 of 10 will be tested positive

Hence out of 60 , only 10 will actually be cocaine uses. It depends only on the proportionality of users.

Hence option C

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When 100 people who have not used cocaine are tested for cocaine use

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