Re: Problem of the Day - 27th May 2014

Posted by Arushi :)) on May 27, 2014; 7:26am
URL: http://discussion-forum.276.s1.nabble.com/Problem-of-the-Day-27th-May-2014-tp7590728p7590732.html

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