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Algebra

Scorching goIITian

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21 Mar 2008 13:28:50 IST

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792

how to get 100% in aieee

None

the title seems different than what i will post

find probability of student scoring 100% in aieee without watching question paper if student knows that 34 % of answer are option a ,36 % of answers are option b ,18 %are of answer are option c and remaining are d
rates are sure for correct answer with their solution

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Akhil

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21 Mar 2008 23:00:26 IST

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good qn..
i am confused, but is the answer
(0.34)³⁴(0.36)³⁶(0.18)¹⁸(.12)¹²

Akhil

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21 Mar 2008 23:05:37 IST

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he answers 34 qns as a, 36 as b, 18 as c, 12 as d
now, if the right answer to a qn is a, there is 0.34 probability that he answers it correctly
for all 34 qns answered correctly, he has to do it 34 times,
ie (0.340(0.34)...34 times ie (0.34)^34
similarly for all options and multiplying , we have

P=(0.34)³⁴(0.36)³⁶(0.18)¹⁸(.12)^{12

it is in the order of 10 raise to -57}

chemicaltester

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22 Mar 2008 17:50:11 IST

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aint the negative mark of aieee is consider in this equation
cos had the same condition been in the entrance examination where no negative marks are given the answer would have been same

Akhil

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22 Mar 2008 17:55:59 IST

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if he gets all qns right anyway, there is no qn of negative marking!

gokul subramanian

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22 Mar 2008 18:25:30 IST

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i think the ans shud be:
{36!*34!*18!*12!}/100!

i no i mentined * 4! previously but after lookin @ ryu's ans i checked my sol n edited it

Akhil

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22 Mar 2008 19:38:58 IST

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plz post ur solution gokul!
i am pretty sure i am wrong

gokul subramanian

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22 Mar 2008 20:32:25 IST

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i want the ans 1st

chemicaltester

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22 Mar 2008 20:58:16 IST

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computer111
agar mere pass answer hota to me yaha pe kyo dalta
kam se kam apna solution to batao

gokul subramanian

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22 Mar 2008 21:02:14 IST

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english pls

RyuAmakusa

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22 Mar 2008 22:12:49 IST

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let there be 100 questions in the paper . now he will mark 34 a's 36 b's 18 c's and 12 d's now only a particular sequence of a,b,c,d,'s is correct and totally there can be

100!

p = --------------------------

34! x 36! x 18! x12!

ways of answering

therefore the required ans is 1/p

chemicaltester

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22 Mar 2008 22:29:42 IST

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ryu
in entrance exm let there be 100 question
in this exm suppose there is no negative marking
then to the probability is 1/p
if there is negative marking then too it is 1/p
i cannot understand this

RyuAmakusa

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22 Mar 2008 22:59:37 IST

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you see this arises because what we want is 100% marks say if we wanted 90% marks then the problem is lot more diff. then we need -ve markin and stuff so that we can get the diff. ways in which he could have got 90%. but since we want 100% marks there is only one case.

thevyzz

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22 Mar 2008 23:10:13 IST

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it depends on the number of questions, RyuAmakusa's answer is correct for 100 questions

the general soln for n questions is p = n! / [ ( 0.34n)! * ( 0.36n)! *( 0.18n)! *( 0.12n)! ]

take reciprocal for probability

RyuAmakusa

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22 Mar 2008 23:28:59 IST

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yeah thevyzz my sol. is only true for 100 questions. & your's is general but i think my explanation for the negative marking is correct.

chemicaltester

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23 Mar 2008 10:00:31 IST

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can u solve this for 90% percent too
if then how?

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