Answer:
B^c and A^c are independent
Step-by-step explanation:
The proof:
We assume that A and B are independent.
By definition, the occurrence of A doesn't affect the probability of B.
Thus, the occurrence of A also doesn't affect the probability of B^c
So by definition, A and B^c are also independent, which by definition again means that the occurrence of B^c doesn't affect the probability of A.
(Here we used the symmetry of independence.)
Therefore, the occurrence of B^c also doesn't affect the probability of A^c.
So by definition, B^cand A^c are also independent.
One could convert the proof to the language of math:
A and B are independent↓P(B|A)=P(B)↓1−P(B^c|A)=1−P(B^c)↓P(B^c|A)=P(B^c)↓A and B^c are independent↓P(A|B^c)=P(A)↓1−P(A^c|B^c)=1−P(A^c)↓P(A^c|B^c)=P(A^c)↓B^c and A^c are independent But now we used conditional probabilities, which might be a problem in case P(A)=0 or P(B^c)=0
OR
Since,
P(A') = 1-P(A)
P(B') = 1- P(B)
Now clearly P(A′)P(B′)=1−[P(A)+P(B)]+P(A∩B)
From set algebra we know that P(A)+P(B)=P(A∪B)+P(A∩B) Substituting, we have P(A′)P(B′)=P([A∪B]′)
Now from De morgans law we know that:
[A∪B]′=[A′∩B′]
Substituting, we have P(A′)P(B′)=P(A′∩B′) , as required.
