let A and B be two events with .... ?

Question

P(B)= .5
P(AUB)= .6
P(A | B)= .4

Find
1- P(AB)
2- P(A)
3- P(A | A)
4- P(B | A)
5- P(A | A) <== there's a line over the last A " A complement "

this is wt i did
1- P(AB)= .5 * .4 = .2
2- P(A)= .6 - .5 + .2 = .3
3- P(A | A) i didnt know how to solve it.
4- P(B | A)= .2 / .3 = .666666
5- P(A | A "complement") didnt know this one too.

Answer 1

Find:

1. P(AB) = P(A | B)*P(B) = .4*.5 = .2

2. P(A)
P(A U B) = P(A) + P(B) - P(AB)
P(A) = P(A U B) - P(B) + P(AB) = .6 - .5 + .2 = .3

3. P(A | A) = 1 by definition
(or you can just plug into the conditional formula in 4.)
P(A) = P(A | A)*P(A)
P(A | A) = P(A) / P(A) = 1

4. P(B | A)
P(AB) = P(B | A)*P(A)
P(B | A) = P(AB) / P(A) = .2/.3 = 2/3

5. P(A | ~A) = 0 by definition

Note: P(A | A) + P(A | ~A) = 1

Answer 2

Your answers look good.
Q 3. They want the probability you are in the non union part of A given that you are originally in A. The answer is 1/3rd
Q5. They want the probability you are any part of A divided by the probality you could be in any part of the space that does not include A (or B) That's like 3/7th

Answer 3

For 3 and 5 just use the formula for conditional probability

P(A|A) = P(A intersect A) / P(A) = P(A) / P(A) =1

and Let A^c= A complement

P(A|A^c) = P(A intersect A^c) / P(A^c)
= P(empty set) / P(A^c) = 0

The rest look fine to me too.