0
$\begingroup$

Given exclusive events $A$, $B$, how is: $P(B) = P(AB) + P(\overline{A}B)$?

I am still confused, how is this true?

  • 0
    When $B$ occurs, either $A$ has also occurred or not. If you write this sentence in mathematical notation, it becomes the one that you mentioned – 2017-01-23
  • 0
    Draw a venn diagram and you will get it yourself – 2017-01-23
  • 2
    $B=$ Shit happens; $A=$ By day, so $\bar A=$ By night. – 2017-01-23

1 Answers 1

0

This is true regardless of whether or not $A$ and $B$ are exclusive.

One of the axioms of probability is that $Pr(E\cup F)=Pr(E)+Pr(F)$ whenever $E$ and $F$ are mutually exclusive.

Note then that $B = (B\setminus A)\cup (B\cap A)$, or using notation more similar to yours $ B=(AB)\cup (\overline{A}B)$

This should be immediately clear, but if its not, notice that $A\cup \overline{A}=\Omega$ where $\Omega$ is the sample space and that $B\cap \Omega=B$ since $B$ is by definition a subset of the sample space. We have then $B=B\cap \Omega=B\cap (A\cup \overline{A})=(B\cap \overline{A})\cup(B\cap A)$

Finally, we notice that $B\setminus A$ is mutually exclusive with $B\cap A$, meaning that we can apply the aforementioned axiom.

$Pr(B)=Pr((B\setminus A)\cup (B\cap A))=Pr(B\setminus A)+Pr(B\cap A)$ or rather, in your notation $Pr(AB)+Pr(\overline{A}B)$