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Looking for a bit of help here. If $B$ and $C$ are disjoint, and $D = B^c \cap C^c$ does that make D=1 and therefore they would form a partition? I think it cannot be that simple. The full question also includes the following. Events $A,B,C,D$ where $P (A\mid B)= P(A\mid C)= P(A\mid D)= 0.5$. $B \cap\ C$ = $\phi\ $ and $D = B^c \cap C^c$
Find P(A).

Not really sure where to start.

fresh143
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  • In the title, $B$ and $C$ are assumed to be zero (by which you mean empty? or maybe probability zero?), but in the body, there is no such assumption. Please edit for consistency. – Gerry Myerson Jun 06 '19 at 13:01
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    Edited sorry, wasn't sure how to use the formatting properly. – fresh143 Jun 06 '19 at 13:10

1 Answers1

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Let $X$ be the whole set, then $B,C$ and $D$ form a disjoint partition of $X$. From this it follows that $$P(A)=P(B)P(A|B)+P(C)P(A|C)+P(D)P(A|D)=\frac{1}{2}(P(B)+P(C)+P(D))=\frac{1}{2}.$$

Floris Claassens
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  • Ah I see. I was trying to figure out P(A)= P(B)P(A|B) + P(A|$B^c)P(B^c)$, not remembering the rule. So if P(A) = 0.5 and P(B+C+D)=1, does that make A a subset of D? – fresh143 Jun 06 '19 at 13:27
  • Definitely not. If $A$ was only a subset of $D$, then $P(A|B)=P(A|C)=0$, as then $$A\cap (B\cup C)\subset D\cap (B\cup C)=(B^{c}\cap C^{c})\cap (B\cup C)=\emptyset$$ – Floris Claassens Jun 06 '19 at 13:36
  • I see my mistake there, with the subset. Just re-reading your original answer, you said that B,C and D form a disjoint partition of X, what then is the relationship between A and D? I'm just trying to picture it as a venn diagram to help me understand. I've never seen the phrase disjoint partition, or is that just a partition where all the elements are disjoint? Sorry for so many questions, and thank you for your quick replies. – fresh143 Jun 06 '19 at 13:44
  • Yes, a disjoint partition is a partition where all elements are disjoint. To give a more practical example of the current situation: Suppose you are throwing a dice, so $X={1,2,3,4,5,6}$ and we take $B={1,2}$, $C={3,4}$ and $D={5,6})$. Then $A={1,3,5}$ satisfies the conditions of this question. i.e. $A\cap B$ is "half the size" of $B$ and the same principal for $C$ and $D$. However I should add that this is reliant on the probability distribution. – Floris Claassens Jun 06 '19 at 13:51