Re: DSE 2011 QUES NO. 36

Posted by Chinni18 on Jun 21, 2012; 5:20am
URL: http://discussion-forum.276.s1.nabble.com/DSE-2011-QUES-NO-36-tp7577919p7578432.html

Aditi, even if there is one ball in the box, it satisfies the "non-empty" restriction.
So the probability is:
P(Chossing 1st Box)*P(Choosing green ball|B1 is chosen) + P(Choosing 2nd box)*P(Choosing green ball|B2 is chosen)
=(0.5)(1) + (0.5)(49/99)
=0.747