# Problem of the Day - 27th May 2014 Classic List Threaded 9 messages Open this post in threaded view
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## Problem of the Day - 27th May 2014

 Suppose 200 students take an exam consisting of M multiple-choice questions, each with N possible answers. All students decide to answer questions randomly and independently, i.e., each student’s probability to answer any question correctly is 1/N, and all the answers are independent. Each question is worth one point, and a score of M − 3 or more is an “A”. What is the probability that at least one student will get an A? Express your answer in terms of M and N. Evaluate this expression for two cases: 20 true-false questions (i.e., M = 20 and N = 2), and 20 questions with 10 options each (i.e., M =20 and N =10.)
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## Re: Problem of the Day - 27th May 2014

 Let X be the number of wrong questions attempted ,, In order to get an A , X is a random variable , which can take values 0, 1,2,3 so for one student the probability of getting an A is P( X=0) + P (X=1)+ P (X=2)+ P (X=3) MC0(1/N)^M + MC1 (1-1/N) (1/N )^M-1 + MC2 (1-1/N )^2 (1/N)^M-2 + MC3 (1-1/N) ^3 (1/N ) M-3 Let this expression be denoted by Y .  the prob of atleast one student getting an A is 1- P( None of them getting an A ) 1- 200C0 ( Prob of getting A)^0 ( prob of not getting an A ) ^200 1- (1- Y )^200  SO for first case : M= 20 and N = 2 we can find the value of Y .. (1/2 )^10 + 20 (1/2 )^20 + 190 (1/2 ) ^20 + 1140 (1/2)^20 (1/2)^20 ( 1+20+ 190+1140 ) (1/2)^20 1351 = .001289 approx= Y  so prob of atleast one of the students getting A is 1- (0.998711)^200
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## Re: Problem of the Day - 27th May 2014

 dunno whether i am right or not .. :'(
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## Re: Problem of the Day - 27th May 2014

 In reply to this post by Prerna Rakheja The probability that a particular student will get Grade A is given by, Pr(Of a getting A for a particular student)= (MC(M-3))*(1/N)^(M-3)*(1/(N-1))^3+(MC(M-2))*(1/N)^(M-2)*(1/(N-1))^2+(MC(M-1))*(1/N)^(M-1)*(1/(N-1))^1+(MCM)*(1/N)^M*(1/(N-1))^0. Now Pr(a particular student does not get A)=1-Pr(He/She gets A). Also Pr(out of 200 students atleast one gets A)=1-Pr(none gets A). Since all events are independent, pr(None gets A)={1-{(MC(M-3))*(1/N)^(M-3)*(1/(N-1))^3+(MC(M-2))*(1/N)^(M-2)*(1/(N-1))^2+(MC(M-1))*(1/N)^(M-1)*(1/(N-1))^1+(MCM)*(1/N)^M*(1/(N-1))^0]}^200. So required probability=1-{1-{(MC(M-3))*(1/N)^(M-3)*(1/(N-1))^3+(MC(M-2))*(1/N)^(M-2)*(1/(N-1))^2+(MC(M-1))*(1/N)^(M-1)*(1/(N-1))^1+(MCM)*(1/N)^M*(1/(N-1))^0]}^200.  "I don't ride side-saddle. I'm as straight as a submarine"
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## Re: Problem of the Day - 27th May 2014

 hi subhayu ,,, i think our methods r same .. jus fr the diff that i have found the proby for getting an A by X as no of wrong ques and u hv done it by the no of ques attempted correctly .. lets take the case of 3 right and rest wrong .. so acc to me : its MC3 *(1-1/N )^3 *(1/N) M-3 n acc to you .. its M C M-3 *(1/N )^M-3*(1/ N-1)^3 all the terms would be same .. except (1-1/N )^3 , acc to u it should be (1/ N-1)^3 please explain this to me Open this post in threaded view
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## Re: Problem of the Day - 27th May 2014

 you are right Arushi, Pr(success)=(1/N), Pr(Failure)=(N-1)/N=(1-(1/N))... It shud be that only..my fault..thanx for pointing it out.  "I don't ride side-saddle. I'm as straight as a submarine"
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## Re: Problem of the Day - 27th May 2014

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