Re: DSE 2011 QUES NO. 36

Posted by sonudelhi on Jun 17, 2012; 9:37am
URL: http://discussion-forum.276.s1.nabble.com/DSE-2011-QUES-NO-36-tp7577919p7577945.html

i did it as.......let first box x red balls and y green balls.then second box would have 50-x and 50-y respectively.

P(G)=P(b1)P(G/b1)+P(b2)p(G/b2)
      =1/2*y/x+y + 1/2*50-y/100-x-y

now maximize P(G) subject to conrs. x+y=1

you will get y=1 and x=0 implies that 0 red and 1 green ball in first box