f two eventsAandBare independent, then we knowP (A∩B) = P (A) P (B). A fact is that ifAandBare independent, then so are all combinations ofA, B, ... etc. Show that if eventsAandBare independent, thenP (A∩B) = P (A) P (B), and thusAandBare independent. (Hint:P (A∩B) = 1-P (A∪B). Then use addition rule and simplify.)

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.