Re: DSE 2012- Question 42-44, 48 and 55-56.

Posted by wolverine on Apr 26, 2013; 10:37pm
URL: http://discussion-forum.276.s1.nabble.com/DSE-2012-Question-42-44-48-and-55-56-tp7580163p7580194.html

.44)    ax+by=0
        Cx+dy =0
       
        P(0)=P(1)=1/2


      we can check the consistency of the above eqn using AX=0
      |A|=ad-bc eqn (1)

       
       a,b,c,d are iid rv
       we will create the sample space and find the joint prob

       a    d      P(A).P(d)
       0    1        1/4
       1    0        1/4
       0    1        1/4
       1    1        1/4
       

       b    c      P(b).P(c)
       0    1        1/4
       1    0        1/4
       0    1        1/4
       1    1        1/4

      from eqn 1 if system of eqn has unique soln ad-bc is nt equal to zero

     case 1 a=0,d=1 gives ad=0...b=1=c gives bc=1  ad-bc nt equal to zero
            so P=1/4.1/4=1/16

     case 2 and 3, when a=1,0 and d= 0,1  resp..ad=0 and b=c=1 gives bc=1...
             p=1/4.1/4+1/4.1/4=2/16


     case 4,5,6    when a=1,d=1..ad=1   b=0,1,0 and c=1,0,1 resp giving bc=0
                   p=1/4.1/4 + 1/4.1/4 +1/4.1/4= 3/16


             adding all the six cases which gives the probability of a unique soln =6/16 =3/8