1prove If A and B are mutually exclusice and A and B are als

1.prove: If A and B are mutually exclusice and A and B are alseo independent , then either A and B has prrobability zero.

2.prove:If A and B are indenpendent then so are A and not B

3.prove:If A and B are indenpendent then so are notA and not B

Solution

1.

A and B are mutually exclusive evens. Thus, P(A and B) = 0

or P(A U B) = P(A) + P(B)

For independent events, P(A and B) = P(A) * P(B)

Thus, IF A and B are independent and mutually exclusive,

P(A and B) = P(A) .P(B) = 0

So, either P(A) = 0 or P(B) = 0.

Hence proved.

2)

(a). Since A and B are independent,

we know that

P(AB) = P(A)·P(B).

Also,

Fact 1: P(B) + P(Bc ) = 1

Fact 2: P(A Bc ) + P(A B) = P(A)

From Fact 1, we have P(A) · P(B) = P(A)[1 P(B^c )]

= P(A) P(A) · P(B^c ).

From Fact 2, we have P(A B) = P(A) P(A B^c ).

Substituting these into the equation P(A B) = P(A) · P(B), we have P(A) P(A B^c ) = P(A) P(A) · P(B^c ).

Subtracting P(A) from both sides of the equation and dividing both sides by 1,

we obtain P(A B^c ) = P(A) · P(B^c ).

3)

The same logic shows that A^c and B^C are independent.

We have analogous key facts for P(A) and P(A^c ):

Fact 1: P(A) + P(A^c) = 1

Fact 2: P(A B^c ) + P(A^c B^c ) = P(Bc )

Therefore, we have P(A)·P(B^c ) = [1P(A^c )]P(B^c ) = P(B^c )P(A^c )·P(B^c ) from Fact 1

and P(AB^c ) = P(B^c ) P(A^c B^c ).

Substituting into the equation P(AB^c ) = P(A)·P(B^c ),

we obtain P(B^c ) P(A^c B^c ) = P(B^c ) P(A^c ) · P(B^c ).

Subtracting P(B^c ) from both sides and dividing by 1,

we obtain P(A ^c B ^c ) = P(A ^c ) · P(B ^c ).

Therefore, Ac and Bc are independent

Hence proved.