question-pt 16, S3, Q 24 (Identify the flaw) Forum

[deebanger]

For this q, through process of elimination, I am able to get to the correct answer. I understand the the flaw here is the premise is talking about likelihood and the conclusion is talking about a number. But I don't get why B is right, and what makes C wrong. Is C a classic lsat trap choice? do the test writers expect us to pick C as it looks good especially after realising what the flaw is (number/proportion).

[Christine (MLSAT)]

Hey deebanger!

Yes, the LSAT totally expects people to see the flaw, realize it's something to do with percents (likelihoods) vs absolute numbers, and just pick (C) for the fact that it mentions percents.

Now, why is (B) right?

The premises tell us that 40+ moms have births that are more likely to be "difficult," and that difficult births are more likely to be ambidextrous babies. This would well support a conclusion that (absent other factors) there should be a higher rate of ambidextrousness with 40+ moms.

For simplicity, let's say the rate of ambidextrousness in difficult births is 80%.

So, 100 young moms, maybe 10 of those are difficult births, 8 of which would be ambidextrous. 100 40+ moms, maybe ALL of those are difficult births, 80 of which would be ambidextrous. In this example, there ARE more ambi kids born to the 40+ moms. But that's only because I kept the number of births for young moms and 40+ moms the same.

Let's change the numbers: now we have 1000 young moms, 100 difficult births, 80 ambi kids. Only 10 40+ moms though, all 10 difficult births, now only 8 ambi kids. In this situation, even though the percent of children born to 40+ moms who are ambidextrous is way higher than the percent of children born to young moms who are ambidextrous, the second pool is so much larger than the first pool that the absolute numbers of ambi-kids are still higher for the young moms.

This is a classic percents vs absolute numbers question. It is identical to this argument:

• PREMISE: Joe spends a higher percentage of his income on Starbucks than Mary does
  CONCLUSION: Joe spends more money at Starbucks than Mary does

This argument overlooks the possibility that maybe Mary makes more than Joe. If Mary made way more than Joe, then she could spend a smaller percentage of her own salary, and yet spend more money overall.

What do you think?

[deebanger]

Hey Christine! thanks so much for your help!, and yes, i get now why B is right, and and for C, is the "percentage of people in the population who are ambi"" irrevalent? is that the only reason is it wrong?

[Christine (MLSAT)]

Hey Christine! thanks so much for your help!, and yes, i get now why B is right, and and for C, is the "percentage of people in the population who are ambi"" irrevalent? is that the only reason is it wrong?

Yep! We don't need to know the percentage of the whole population that is ambidextrous - the argument doesn't assume anything about that percentage to achieve the conclusion. So, the argument DOES "fail to specify" that, but that's not a flaw!

The central assumption the argument has to be making is about the relative size of the two populations of kids (those of young moms and those of 40+ moms).

Make sense?


PS - did the explanation for the other question you posted this morning make sense? I want to be sure, as I know that one was stickier!