Re: Problem of the Day - 27th May 2014

Posted by Prerna Rakheja on May 27, 2014; 6:18pm
URL: http://discussion-forum.276.s1.nabble.com/Problem-of-the-Day-27th-May-2014-tp7590728p7590832.html

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