Conditional probability with union evidence 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. 

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^+) .      } 
}$$
