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Here goes... Let's say you take the Quant section of the

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mciatto
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Here goes... Let's say you take the Quant section of the [#permalink] New post   14 Aug 2003, 08:59
Here goes...

Let's say you take the Quant section of the GMAT and get 33/37 correct. Now after the test is done, you find out that 5 of the 37 are "trial questions" that do not count. What are the chances that you got a perfect score on this section?



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Re: Here goes... Let's say you take the Quant section of the [#permalink] New post   14 Aug 2003, 23:35
A cool problem. I have thought half an hour what to do.
You score perfect if all your wrong answers (4) are not counted (5).
In other words, for 5 noncounted units, 4 should be wrong plus one correct.

favorable outcomes: 4C4 (4 wrong answers)*33С1 (1 correct)
total 37C5

33/37C5=33*32!*5!/37!=1/17*7*3*37=1/13209

a sanity check: the smallest probability to score perfect is the case when 4 questions are not counted, we have 5, so the probability should be very small

When such a probability is any seizable? Let us imagine that 15 questions are not counted. Answering 33/37, we should have far more chances to score perfect. 4C4*33C11/37C15=0.02

comments?
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Re: Here goes... Let's say you take the Quant section of the [#permalink] New post   14 Aug 2003, 23:49
stolyar wrote:
A cool problem. I have thought half an hour what to do.
You score perfect if all your wrong answers (4) are not counted (5).
In other words, for 5 noncounted units, 4 should be wrong plus one correct.

favorable outcomes: 4C4 (4 wrong answers)*33С1 (1 correct)
total 37C5

33/37C5=33*32!*5!/37!=1/17*7*3*37=1/13209

a sanity check: the smallest probability to score perfect is the case when 4 questions are not counted, we have 5, so the probability should be very small

When such a probability is any seizable? Let us imagine that 15 questions are not counted. Answering 33/37, we should have far more chances to score perfect. 4C4*33C11/37C15=0.02

comments?


This question is a little ambiguous. Your answer assumes that it is equally likely that you will miss an experimental question as a real one. But is that really true? I would think that the experimental question are somewhat more difficult, hence ETS needs to test them to determine whether they are fair, unambiguous, solvable, etc.... If the chance that you will miss an experimental question is significantly higher than a real one, it would raise the probability of increasing your score. However, since we have no data regarding the relative difficulty of real vs. experimental questions, I don't think you can calculate an answer.
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Re: Here goes... Let's say you take the Quant section of the [#permalink] New post   15 Aug 2003, 00:17
I felt the same ambiguity. Yet, I hope the approach is relevant, at least partly.
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Re: Here goes... Let's say you take the Quant section of the [#permalink] New post   15 Aug 2003, 01:30
stolyar wrote:
I felt the same ambiguity. Yet, I hope the approach is relevant, at least partly.


As I stated, your approach is exactly correct under the given assumption. However, bad assumption is almost always equal to bad result.
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Re: Here goes... Let's say you take the Quant section of the [#permalink] New post   19 Aug 2003, 21:14
Herez my 2 cents worth. I think the probability of getting a perfect score is 25%.
Reason? Total chances for scoring 33/37

Excuse my abbreviations
OTCS -- Outside trial correct score
TCS -- Trial correct score x/5
TWS -- Trial wrong score y/5
OTWS -- Outside trial wrong score

OTCS -- TCS -- TWS -- OTWS
1) 32--1--4--0
2) 31--1--3--1
3) 30--1--2--2
4) 29--1--3--4

So there are 4 possibilities that YOUr score is distrubuted among the Trial and non-Trials.

Case 1 only yieds a perfect score. Meaning no one can score above 32/37.

Therefore the chances of YOU getting a perfect score despite the trial questions are 25%.

Let me know if I was ignorant of any fact that was obvious.
Thanks
Chaitram
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Re: Here goes... Let's say you take the Quant section of the [#permalink] New post   20 Aug 2003, 08:16
chaitramasam wrote:
Herez my 2 cents worth. I think the probability of getting a perfect score is 25%.
Reason? Total chances for scoring 33/37

Excuse my abbreviations
OTCS -- Outside trial correct score
TCS -- Trial correct score x/5
TWS -- Trial wrong score y/5
OTWS -- Outside trial wrong score

OTCS -- TCS -- TWS -- OTWS
1) 32--1--4--0
2) 31--1--3--1
3) 30--1--2--2
4) 29--1--3--4

So there are 4 possibilities that YOUr score is distrubuted among the Trial and non-Trials.

Case 1 only yieds a perfect score. Meaning no one can score above 32/37.

Therefore the chances of YOU getting a perfect score despite the trial questions are 25%.

Let me know if I was ignorant of any fact that was obvious.
Thanks
Chaitram

My take on this (AkBrah, let's assume all prob is equal)..

There are 37C5 ways that the "trial questions" can be distributed.

33C1 of them will include the four that you got wrong.

Prob is 33C1/37C5.
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Re: Here goes... Let's say you take the Quant section of the [#permalink] New post   20 Aug 2003, 23:45
As mentioned by AkamaiBrah, the difficulty level of the trial question will influence the outcome of answering a trial question right, in reality. Ignoring this fact for a moment,

We have 33 correct answers out of 37. 4 incorrect answers. 5 of those 37 questions are trial questions.

We can approach the problem in two ways, first (as stolyar approached),

33 correct answers, 4 wrong answers - choose 5 trial questions among them. Lets find out the probability of all 4 wrong answers being that of the trial questions.

p = (4 wrong being trial * one correct being trial )/ (choosing 5 trial from 37)
p = 4C4 * 33C1/37C5 = 1 *33/435893 = 1/13209

Second approach is,

32 real questions, 5 trial questions; find the probability of 4 wrong answers being that of the trial questions. [32 red balls, 5 blue balls - find the probability of 4 balls chosen being blue]

p = 5C4/37C4 = 5/66045 = 1/13209
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Re: Here goes... Let's say you take the Quant section of the [#permalink] New post   21 Aug 2003, 01:56
bnagasub wrote:
As mentioned by AkamaiBrah, the difficulty level of the trial question will influence the outcome of answering a trial question right, in reality. Ignoring this fact for a moment,

We have 33 correct answers out of 37. 4 incorrect answers. 5 of those 37 questions are trial questions.

We can approach the problem in two ways, first (as stolyar approached),

33 correct answers, 4 wrong answers - choose 5 trial questions among them. Lets find out the probability of all 4 wrong answers being that of the trial questions.

p = (4 wrong being trial * one correct being trial )/ (choosing 5 trial from 37)
p = 4C4 * 33C1/37C5 = 1 *33/435893 = 1/13209

Second approach is,

32 real questions, 5 trial questions; find the probability of 4 wrong answers being that of the trial questions. [32 red balls, 5 blue balls - find the probability of 4 balls chosen being blue]

p = 5C4/37C4 = 5/66045 = 1/13209


Consider this simple approach:

He got four wrong. The chances of him getting any particular four questions wrong are the same so we can focus on just these four questions. He needs to match all four wrong answers with experimental questions, of which there are 5 in 37.

p = 5/37 * 4/36 * 3/35 * 2/34 = 1/13209
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Re: Here goes... Let's say you take the Quant section of the [#permalink] New post   21 Aug 2003, 06:24
mciatto wrote:
My take on this (AkBrah, let's assume all prob is equal)..

There are 37C5 ways that the "trial questions" can be distributed.

33C1 of them will include the four that you got wrong.

Prob is 33C1/37C5.


Just for the record, 33C1/37C5 = 1/13209.

Love finding more than one way to do a problem!! :-D



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Re: Here goes... Let's say you take the Quant section of the [#permalink]
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