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.

maximum area = xy = (16)(32) = 512 ft2

to maximize the area of the garden, the side of the fence parallel to the side of the house should be 32 ft long, and the other two sides should both have length 16 ft.'

source

the terms of a geometric sequence can be found by beginning with the first term and multiplying by the common ratio repeatedly. a recursive formula for a geometric sequence with common ratio r is given by an=ran−1 a n = r a n − 1 for n≥2 n ≥ 2 .

step-by-step explanation: