ISI 2017 key

[Asd1995]

It wasn't given that p is a random variable, either..

[Econ17]

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[Halflife]

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[Econenthusiast]

Yes, I assumed that it was a pdf right away after looking at the question... shouldn't have done that

[Asd1995]

Happens man, I screwed up a couple of straightforward ones too.

[Econenthusiast]

Yeah me too. Didn't expect the paper to go this bad

[Econ17]

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[Halflife]

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[Asd1995]

Yeah I can verify from a relatively reliable source the cutoff was 22/30 for Math.

Any idea what the experts have to say about this year's PEB?

[Halflife]

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[abhi.roop18]

Hey guys..Is anyone interested in discussing the answers to Pea and peb?

[Asd1995]

Woah that's pretty low.. Guess I'm pretty much screwed coming from a math background lol

[Asd1995]

There's a conjecture floating around that this year cutoffs are not individual. Can someone verify?

[Halflife]

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[ElPsyCongroo]

Monkey question is related to random walk, you can google it online.

The second part I did not get but the first part I got to be 1 by elimination.

Basically I thought about the ways in which pi_1 could be true (that is, the monkey reaching 1 at some t>0).

So, that could be p (monkey moves right only) (1-p)^2(p)^3 (monkey moves two left and three right) and after this point I thought about a lot of other stuff but then that was not relevant to the answer. The moment you know pi_1 > p + (1-p)^2(p)^3, you can eliminate all options except the one for pi_k=1

Given p is greater than half and less than 1, then pi_1 can't be p^k, since that could imply pi_1 is p, but we know it's greater than p, ditto for the other option (I guess that was p/k) and for the option that said (p/(1-p))^k you can eliminate straight up because if p>1/2 this becomes  more than 1, and check the case for p=1/2 that will also give you something not possible.

Can't be zero also since pi_1 > p

So, it must be 1. There is a more formal solution, I found a pdf once, I'll post it here later if I find.

[ElPsyCongroo]

Also, I believe I attempted about 23 in PEA and I think a little more in PEB.

[Econenthusiast]

Thanks EIPsyCongroo. Could you, or anyone else share how you did that question on raffle tickets if in case you had attempted that ?

[ElPsyCongroo]

The one with Mr. A no?

So p(Sell to all) = 1-p(Not sell to all)

It's possible to solve it from this point because if you think about it, the only way he won't be able to sell tickets to everyone is if the 2-rupee guy is first in line, (otherwise he'll have enough 1 rupee coins with him to give the guy change).

So, the probability of the 2rs. guy being first is (number of ways 2rs. guy can be first)/(total number of arrangements).

So that's (n-1)!/(n)!=1/n

So answer is 1-1/n = (n-1)/n, I think that should be right.

[Econenthusiast]

Yes I guess that is right. Thanks a lot for your solution!!

[Akshay Pamnani]

Hi guys,
Do you think 85 in both papers should do it?
I am not able to get the exam out of my mind.
I took maths too lightly,coz I was able to solve atleast 25-28 questions in previous papers,and the fact that I am from IIT,I thought it would not be an issue.
Did not expect so many questions from Statistics and college level maths(Did prepare it,but not in depth),think I have totally messed up the maths paper,coz of that.

Just keeping my fingers crossed.