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.

87%

step-by-step explanation:

apex answer

step-by-step explanation:

let

we know that

the y-intercept is the y-coordinate of the point when the value of x is equal to zero

so

the y-coordinate of the point a is the y-intercept

the answer part a) is

the value of y-intercept is

part b) find the slope m

we know that

the value of the slope is equal to

substitute the values

therefore

the answer part b) is

the value of the slope is equal to

the general answer is

the slope is

the value of y-intercept is