I am having trouble answering the following question

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


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


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

  • $\begingroup$ 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 $A\cup B$. $\endgroup$ – André Nicolas Sep 6 '14 at 20:28
  • $\begingroup$ Have you tried drawing a Venn diagram? In this simple case, the answers will pop out at you. $\endgroup$ – Dilip Sarwate Sep 6 '14 at 20:28
  • $\begingroup$ 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. $\endgroup$ – Fernando Martinez Sep 6 '14 at 20:38
  • $\begingroup$ 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? $\endgroup$ – André Nicolas Sep 6 '14 at 20:42
  • $\begingroup$ So I guess it would be .37+.56=.93 $\endgroup$ – Fernando Martinez Sep 6 '14 at 20:47

Look at this diagram: enter image description here $$$$Could you find your answers?


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