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In the problem of cancer (C) and tests (t1, t2), or any other example,

How can I calculate: $P(C^+|(t1^+ \text{ or } t2^+)$ I think this would be the same as finding: $P(t1^+ \text{ or } t2^+|C^+) P(C^+)\over P(t1^+ \text{ or } t2^+).$

But is $P(t1^+ \text{ or } t2^+|C^+) = P(t1^+|C^+)+P(t2^+|C^+)-P(t1^+ \text{ and } t2^+|C^+)?$

On the other side, is it true in other problems that $P(t2^+|t1^+) ={ P(t1^+ \text{ and } t2^+)\over P(t1^+)}?$

Thanks.

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The answer for your third (last) question is "yes"; this is just the definition of conditional probability. (I answer this first, since it is used later, here).

Your initial instinct is right. For any two events $A$ and $C$: $ P(A|C)={P(C\cap A)\over P(C)}={P(A\cap C)\over P(C)}={P(A)P(C| A)\over P(C)}. $

The answer to your second question is "yes": $\eqalign{P(t1^+\cup t2^+|C^+)&={P(( t1^+\cup t2^+)\cap C+)\over P(C^+)}\cr & ={P( ( t1^+\cap C^+)\cup (t2^+\cap C^+))\over P(C^+)}\cr &={P( t1^+\cap C^+)+P (t2^+\cap C^+)- P (t2^+\cap t1^+\cap C^+) \over P(C^+) }\cr &={P( t1^+\cap C^+) \over P(C^+) } +{P (t2^+\cap C^+) \over P(C^+) } -{ P (t2^+\cap t1^+\cap C^+) \over P(C^+) }\cr &={P( t1^+| C^+)+P (t2^+| C^+)- P (t2^+\cap t1^+| C^+) . } }$