Probability of events expressed in terms of two mutually exclusive events

I am having trouble answering the following question

If $A$ and $B$ are mutually exclusive events, $P(A)=.37$, and $P(B)=.44$:

Find

a. $P(A^C)=1-.37$

b.$P(B^C=1-.44$

c.$P(A\cup B)=.37+.44=.81$

But I am having problem with this part

D.$P(A \cap B)=0$ I think this is zero because they are mutually exclusive

E. $P(A \cap B^C)=$ I am not sure how to this one

F. $P(A^C \cap B^C)$ or this one

• On D, you are right. For E, $A$ and $B$ are mutually exclusive, so $A$ is a subset of $B^c$. Thus the required probability is $\Pr(A)$. I will at least for a while leave the last one to you. Draw a picture (Venn diagram) and it may become clear. Or if you like algebraic methods, note that the complement of $(A^c\cap B^c)$ is $A\cup B$. – André Nicolas Sep 6 '14 at 20:28
• Have you tried drawing a Venn diagram? In this simple case, the answers will pop out at you. – Dilip Sarwate Sep 6 '14 at 20:28
• I get a bit confused about what and is in mutually exclusive. Or is add up the two probabilities. And is their intersection. So A=.37 B^C=.56 so is AND subtract. – Fernando Martinez Sep 6 '14 at 20:38
• Think of probability as weight. If two regions do not overlap and one has weight $x$ and the other has weight $y$, what is the combined weight of the two regions? – André Nicolas Sep 6 '14 at 20:42
• So I guess it would be .37+.56=.93 – Fernando Martinez Sep 6 '14 at 20:47

Look at this diagram: Could you find your answers?