The events A, B, C have the probabilities: P(A|B)=0.25, P(C|B)=0.5, P(A∩C|B)=0.10. Given that B has happened, find the following probabilities:
a) That only C has happened 
b) That only C or only A has happened, but not both of them
c) That C or A has happened
I have answered the three questions, but want to know if the logic is correct (A' = A compliment):
a) P(A'∩C|B) = P(C|B) - P(A∩C|B) = 0.5 - 0.1 = 0.4
On the RHS I simply take the probability that C happened given B, and subtract from it the probability that A and C happened. The LHS is the only way I can find of representing only C without including C∩A, though I'm not sure if this is the correct way of representing that.
b) P(C∪A|B) = P(C|B) + P(A|B) - 2*P(C∩A|B) = 0.5 + 0.25 - 0.2 = 0.55
Addition rule of probabilities. Subtract twice the intersection, once for repeat, and second because we don't want to include it at all.
c) P(C∪A|B) = P(C|B) + P(A|B) - P(C∩A|B) = 0.5 + 0.25 - 0.1 = 0.65
Same as above, except only subtract intersection once.
Is my logic above correct?
a) if by "only C has happened" it means among the two, then you are correct. However, if it means among the three, then the probability is zero $$P(A'\cap B' \cap C|B)=0$$
b) $$\begin{align}P(\text{only A or C}|B)&=P((A\cap C')\cup(A'\cap C)|B)\\ &=P(A\cap C'|B)+P(A'\cap C|B)-P((A\cap C')\cap(A'\cap C)|B)\\ &=P(A|B)-P(A\cap C|B)+P(C|B)-P(A\cap C|B)-0 \end{align}$$
c) is correct