Problem of the Day - 27th May 2014

[Prerna Rakheja]

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.)

[Arushi :))]

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

[Arushi :))]

dunno whether i am right or not .. :'(

[Granpa Simpson]

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.

[Arushi :))]

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

[Granpa Simpson]

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.

[Nupur]

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[Nupur]

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[Prerna Rakheja]

P(at least one student gets A) = 1 − P(every student has at least 4 wrong answers) P(every student has at least 4 wrong answers) = (P(one student has at least 4 wrong answers))^200 Because the students answer independently, the probability that every student has at least four incorrect answers is equal to the product, over all 200 students. P(one student has at least 4 wrong answers) = 1 − P(one student has at most 3 wrong answers) The probability for a student to get exactly one answer incorrectly is M ·(1/N)^(M−1)·(1 − 1/N) The number of ways to get two incorrect answers out of M is = M(M − 1)/2 The number of ways to get three incorrect answers out of M is = M(M − 1)(M − 2)/6 Therefore, the probability for a student to get at most three incorrect answers is = ∑_(k=0)^3〖MCk.(1/N)^(M-k) ((N-1)/N)^k 〗 = 1-[1 -(1+M.(N-1) + (M(M-1) (N-1)^2)/2+ M(M-1)(M-2)(N-1)^3)/6) ].(1/N)^M}^200 For M = 20 and N = 2 it simplifies to Pr(Atleast one student gets an A) = 1 - (1 - 1315(1/2)^20)^200 approx = 0.227 You can similarly solve for N = 10