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.

step-by-step explanation: the answer for wha if rt is 10 centimeters long, and wha is st? it’s 6