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

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.

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.

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

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

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