Question

If A and B are two independent events, prove that A' and B are also independent.

Answer 1

Solution:

We know that if two events A and B are independent than

p(A intersection B) = p(A).p(B) ...eq1

and we know that

p(A' intersection B') =1- p(A U B)

As p(A U B)= p(A)+p(B)-p(A intersection B)

So

p(A' intersection B') =1- p(A)-p(B)+p(A intersection B)

=1- p(A)-p(B)+p(A).p(B) from eq1

=[1-p(A)]-p(B)[1-p(A)]

=[1-p(A)][1-p(B)]

p(A' intersection B')=p(A').p(B') ...eq2

since
[1-p(A)]= p(A')

[1-p(B)]=p(B')

hence we can check from eq1 and eq2 that if A and B are two independent events than A' and B' are also independent.

Hope it helps you.

Note: don't find symbol of intersection,so write instead

Answer 2

Solution:

Concept:

If two events A and B are mutually exclusive events then

P(A∪B) = P(A)+P(B)

If two events A and B are independent

if P(A∩B) = P(A).P(B)

Given:

A and B are independent

Then,

P(A∩B) = P(A).P(B)

since A∩B and A'∩B are mutually exclusive events and B= [A∩B]∪[A'∩B],

we have

P(B)=P(A∩B)+P(A'∩B)

This implies

P(A'∩B) = P(B) - P(A∩B)

P(A'∩B) = P(B) - P(A).P(B)

P(A'∩B) = P(B).[1 - P(A)]

P(A'∩B) = P(B).P(A')

P(A'∩B) = P(A').P(B)

Hence A' and B are independent events.