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I was finishing up my statistics homework But I was unsure if I was thinking of the last problem correctly.

It reads

If $P(ABC)=0.2$, are $A$ and $C$ mutually exclusive?

My thinking is that mutually exclusive means $A\cap C = \emptyset$ And if that's true $P(ABC)$ would equal $0$, so they are not mutually exclusive.

Am I thinking about this correctly? If not could someone please help me with how to do this question?

Thank you.

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  • $\begingroup$ What does the symbol "B" mean? $\endgroup$ – epimorphic Jan 20 '16 at 1:13
  • $\begingroup$ @epimorphic It's just another event, so events $A,B,C$. $\endgroup$ – Em. Jan 20 '16 at 1:14
  • $\begingroup$ Formatting tips here. $\endgroup$ – Em. Jan 20 '16 at 1:34
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Clearly $ABC\neq \varnothing$. So there exists $x\in ABC$, which means $x$ is a member of $A$ and $B$ and $C$, but if it is a member of $A$ and $C$, then it is a member of $AC$. Therefore $AC\neq \varnothing$, which means $A$ and $C$ are not mutually exclusive.


I think you can also do what you're saying, if you're allowed to know that intersection is commmutative. So, what I mean is, let $P(ABC) = .2$, and suppose $AC=\varnothing$. Then $AC\cap B = \varnothing$. But this gives that $$P(ABC) = P(\varnothing) = 0,$$ which is a contradiction since we know that $P(ABC)=.2$. Thus, it must be the case that $AC\neq \varnothing$. So, $A$ and $C$ are not mutually exclusive.

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  • $\begingroup$ Thanks, that is a much more justified explanation. I appreciate your editing of the question as well. I will look more into the sites text formatting in the future. $\endgroup$ – Howmuchmore Jan 20 '16 at 1:31
  • $\begingroup$ @Howmuchmore Thanks for reminding me, I usually leave that as a comment. Also, no problem. For future reference, if you find answers acceptable or correct, then consider giving clicking the check mark next to the answer. Good luck. $\endgroup$ – Em. Jan 20 '16 at 1:34
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$ABC \subset AC \implies P(ABC) \leq P(AC)$. Thus, $P(AC) \geq 0.2$ and so $A$ and $C$ are not mutually exclusive.

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  • $\begingroup$ Great answer. (+1)! $\endgroup$ – user98186 Jan 20 '16 at 20:28

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