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Let $A$ and $B$ denote two events, and suppose that

$\text{P}(A) = 0.2$, $\text{P}(B) = 0.3$, and $\text{P}(A\cap B) = 0.1$.

Are the following computations correct?

  1. $\rm{P}(A \cup B) = \rm{P}(A) + \rm{P}(B) - \rm{P}(A\cap B) = 0.4$
  2. $\rm{P}(B') = 1 - \rm{P}(B) = 0.7$, where $B'$ denotes the complement of $B$
  3. $\rm{P}(B\cap A') = \rm{P}(B)\times \rm{P}(A') = 0.24$
  4. $\rm{P}(A' \cup B') = \rm{P}(A') + \rm{P}(B') - \rm{P}(A'\cap B') = 0.94$

[NB: $\cup$ = "or" and $\cap$ = "and"]

Comp_Warrior
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thedutchman
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    By assuming that $P(B\cap A')=P(B)P(A')$, you are claiming that $A$ and $B$ are independent. Is that the case? – assumednormal Jul 16 '13 at 00:30
  • All that is given is what I wrote there. – thedutchman Jul 16 '13 at 00:33
  • Ok. If $A$ and $B$ were independent, then $0.1=P(A\cap B)=P(A)P(B)=.06$. That's clearly not the case. What if you wrote $P(B)=P(B\cap A)+P(B\cap A')$. Does that help you determine $P(B\cap A')$? – assumednormal Jul 16 '13 at 00:36
  • Since we're given P(AB) as 0.1 and P(B) as 0.3, I could just rearrange the equation and obtain 0.2 as P(BA'). Which is close to what I got (0.24). Is my method of using P(B)*P(A') incorrect then? – thedutchman Jul 16 '13 at 00:40
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    Unfortunately there is no "close" here. Either you are correct or not. – assumednormal Jul 16 '13 at 00:43
  • What I don't get is that if I multiply P(A) and P(B) I get 0.06, yet it is listed as 0.1. What could that mean about the probabilities? This explains the variations between your answer and mine as well. If I use the value of 0.06 for P(BA) I get 0.24 with your equation. – thedutchman Jul 16 '13 at 00:47
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    All this means is that your events $A$ and $B$ are not independent. So knowing that $A$ has occurred tells you something about the probability of $B$ occurring, and vice versa. – assumednormal Jul 16 '13 at 00:49
  • As they are not independent, P(BA') would then be P(B)*P(A' | B) I believe. Thanks for the help :) – thedutchman Jul 16 '13 at 00:57
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    Exactly. For problems with a small number of events, I like to draw venn diagrams to figure out what I know and what I'm looking for. – assumednormal Jul 16 '13 at 01:58
  • Max, you may find the closely related thread at http://stats.stackexchange.com/questions/47671/what-is-the-difference-between-using-the-multiplication-rule-or-using-venn-diagr provides the information you are looking for. – whuber Jul 16 '13 at 14:49

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