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There are three events, $A$, $B$ and $D$. I know that $P(D)=0.2$, $P(A)=0.34$ and $P(B)=0.43$. I have calculated that $P(D\mid A)=0.5294$ and $P(D\mid B)=0.44186$. Now I need to calculate $P(D\mid (A\cap B))$, and I am stuck! I have tried expanding using Bayes' Theorem and I have had no success. One idea I had was that maybe $P(D\mid (A\cap B))=P(D\mid A)P(D\mid B)$ but I don't think this is generally true.

I also know that $P(A^c\mid D)=0.1$, $P(B^c\mid D)=0.05$, $P(A^c\mid D^c)=0.8$ and $P(B^c\mid D^c)=0.7$.

How can I go about calculating $P(D\mid (A\cap B))$ using what I have above?

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1 Answer 1

up vote 3 down vote accepted

I don't think you can find a fixed number for $P(D|(A\cap B))$ but you should be able to put bounds on it.

Since $P(A \cap D)=0.18$ and $P(B \cap D)=0.19$, you have $P(B^c \cap D) = 0.01$ and so $0.17 \le P(A \cap B \cap D) \le 0.18$, and $P(A \cap B \cap D^c) \le P(A \cap D^c) =0.16$. So $$ \frac{17}{33} \le P(D|(A\cap B)) \le 1$$ and these are tight bounds as shown below

enter image description here

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Thanks. I was spending a lot of time trying to calculate it exactly! –  user64219 Nov 19 '12 at 16:48
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