Computation of conditional probabilities of composite events

Question

Let $P(A) = 0.5, P(B|A) = 0.2$ and $P(A \cup B) = 0.8$

Find

a) $P(A \cap B)$
b) $P(A | B^c)$
c) $P(A \cap B^c | A \cup B)$

So for

(a) I know that P( B|A ) = $P(A \cap B)$ / P(A). So $P(A \cap B)$ = .5 * .2 = .1.

I'm confused as to what (b) is asking for. The probability of A given the complement of B, so I think the complement of B is 0.7 but I'm still confused as to how I'm supposed to approach this problem. Same with (c)

Answer 1

Hint$$p(B/A)=\frac{p(A\cap B)}{p(A)}\implies p(A\cap B)=\cdots\\ p(A\cup B)=p(A)+p(B)-p(A\cap B)\implies p(B)=\cdots\\ p(A/B^c)=\frac{p(A\cap B^c)}{p(B^c)}=\frac{p(A)-p(A\cap B)}{1-p(B)}=\cdots$$

Answer 2

Well, in each case, you're just calculating a conditional probability. For part (b), you'll need to find $P(A\cap B^c)$ and $P(B^c),$ first. The result from part (a) will be essential, there. (Hint: How are the probabilities of $A,B,A\cap B$ and $A\cup B$ related?)

For part (c), you'll need to find the probability of $(A\cap B^c)\cap(A\cup B).$ But wait! That's just $A\cap B^c,$ so we've already found it!

Answer 3

You should use a diagram tree and then fill each node with the values you have.