Dse 2012 q.43 44

[vandita24x7]

[Sris]

44...

there will be 16 cases here...since  alpha beta emu and v can take 2 - 2 (0 and 1)values each...so 2*2*2*2 =16 cases...some of them i m writing below...
1 1    1 1   1 1   1 0   0 1   1  0   0  1   0 1   1 0  0  0   0  1   1  0  0 0   0  0  0 0   1 1
1 1    1 0   0 1   1 1   1 1   1  0   0  1   1 0   0 1  0  1   0  0   0  0  1 0   0  0  1 1   0 0

now since it is a homogenous system it will have at least one solution(it cant be inconsistent) so statement b is true

now homogenous system has a unique solution only if det A IS NOT EQUAL TO ZERO...just find the determinants of the above 16 2*2 matrices and see the cases where det A is not equal to zero...that will answer ur statement a....

it will come out to be true ...so both are correct

[Sris]

 there are 10 elements

P(A)=4/10  

Now B can have maximum of 4 and min of 1 element 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


courtesy: kangkan

:)

[bear]

hey sris can u please elaborate on Q43?

[Sris]

what have u not understood

[amianand]

Consider 'p' outcomes that occur in both events, while 'q' outcomes in event B.
Now, For A & B to be independent
P(A ∩ B) = P(A).P(B)
(p/10) = (4/10).(q/10)
so, p/q = 2/5
while q being positive natural number <= 10, hence it can only be 5 or 10.