Re: DSE 2012 paper discussion-Please join

Posted by kangkan on Jun 14, 2014; 7:47pm
URL: http://discussion-forum.276.s1.nabble.com/DSE-2012-paper-discussion-Please-join-tp7590218p7592744.html

43... there are 10 elemetsn

P(A)=4/10  

Now B can have maximum of 4 and min of 1 elemnt common with A

If A and B are inde,than P(A inter B)=p(a)*p(b)

If no of elemts common is 1,then it will imply p(b)=.25 ..but 0.25 is an inadmissible value since p(b) can only take values 0.1,0.2..,0.9 ,1.0

no of coomon elemts 2 P(b)=0.5(admissible)

No of coomon elemts =3, p9b)=0.75 (inadmssible)

No of common elemts =4 p(b)=1 (admissible)

So p(b)=0.5 or 1.0 hence no of elemts 5 or 10

:)