2+2 is always 4. cos2x + sin2x is always 1.
However, this is not always the case. Mathematics also deals with probability - a deterministic science.
Why Guess?
Objective examinations are considered a test of analytical skills and a test of nerves.
Now, it also tests your knowledge of probability. (Didn't it always test that?)
Regardless of the syllabus, objective pattern exams will test how well you apply probability.
Surprised?
Consider this. JEE in 2008 had 132 questions. In these, I would reckon there were 20 questions where I had a doubt. I wasn't sure. It was 50-50.
Did I guess?
Of course, I did.
Guessing in JEE
The objective questions had a +4/-1 marking pattern (+4 if you get it right, -1 if you get it wrong).
I figured out - even if I got 4 questions right out of 20, I wouldn't be losing out. (4x4 - 16x1 = 0).
How much probable is that?
Probability of getting exactly no question right = 20C0(1/2)20
Probability of getting exactly one question right = 20C1(1/2)20
Probability of getting exactly two questions right = 20C2(1/2)20Probability of getting exactly three questions right = 20C3(1/2)20
Probability of getting exactly three questions right = (1/2)20 { 20C0 + ... + 20C3} = (1/2)20 (1351) = 1.28 x 10-3
Probability of getting at least 4 right = 1 - 1.28 x 10-3 ≈ 1
Once I figured this out, my JEE paper was practically finished.
I guessed left-right and finished with a rank of 1594.
Double Check
Still cynical?
Feel that something somewhere is wrong.
Let us double check.
Probability of getting either none right or one right or... 20 right = (1/2)20 { 20C0 + ... + 20C20}
How much is 20C0+20C1+...+20C19+20C20?
It is 220!
We then get the probability = 220 / 220 = 1
That is what we should be getting!
What if I had no idea and simply guessed?
Probability of getting exactly
none right would then become 20C0(3/4)20
one right=20C1(3/4)19(1/4)1
two right = 20C2(3/4)18(1/4)2
three right = 20C3(3/4)17(1/4)3
Probability of getting less than 3 right = (3/4)17 { 20C0(3/4)3 + 20C1(3/4)2 (1/4)1 + 20C2(3/4)1 (1/4)2 + 20C3(1/4)3 } = 0.225
Probability of getting at least 4 right now becomes 1 - 0.225 = 0.775
It isn't as safe to guess now, even though the probability is still in the candidate's favour.
AIEEE/BITSAT etc.
AIEEE has the +3/-1 marking scheme.
If you use the same method to calculate, you'll find that
the probability of scoring non-negative in case of 50-50 ≈ 1
when one has no idea = 0.61
BITSAT has bonus questions if you complete all the questions. Suppose you have 20 questions left out. Should you guess and go for the bonus questions?
You are now empowered to decide for yourselves!
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