If A and B are indepebent show that A and B are also indepe

If A and B are indepebent show that A \' and B\' are also independent

Solution

if A and B are independent than we know that,

P(A) + P(A\') = 1 -------------------1)

and,

P(A B\') + P(A\' B\') = P(B\')

P(A B\') = P(B\')P(A\' B\'). -----------------------2)

and we need to prove that P(A\' B\') = P(A\') · P(B\').

also we know that,

P(B) + P(B\') = 1

P(A B\') + P(A B) = P(A) so,

P(A B) = P(A) - P(A B\') --------------------3)

as A and B are independent

P(A B) = P(A) P(B) --------------4)

but,

P(A) · P(B) = P(A)[1 P(B\')] = P(A) P(A) · P(B\') ----------5)

from 3) ,4) ,5) we have,

P(A) - P(A B\') = P(A) P(A) · P(B\')

so,

P(A B\') = P(A) · P(B\') ---------------6)

from 1) and 2) we have,

P(A) + P(A\') = 1   

P(A B\') = P(B\')P(A\' B\') -----------7)

from 6)  P(A B\') = P(A) · P(B\')

so,

P(A) · P(B\') =  P(B\')P(A\' B\')

P(B\')P(A\' B\') = P(A) · P(B\')

but ,

P(A) · P(B\') =  [1P(A\')]P(B\') = P(B\')P(A\')·P(B\')

so,

P(B\')P(A\' B\') =  P(B\')P(A\')·P(B\')

-P(A\' B\') = P(A\')·P(B\')

P(A\' B\') = P(A\')·P(B\')

so we have prove that  P(A\' B\') = P(A\') · P(B\')

hnece A\' and B\' are independent.