Question 25 of Feller's "Introduction to Probability Theory and Its Applications" in chapter 6 asks the following (page 171):
Multiple Bernoulli Trials. In example (9.c) find the conditional probabilities p and q of (S, F) and (F, S), respectively, assuming that one of these combinations has occurred...
The context from example (9.c) is:
Two sequences of Bernoulli trials with probabilities of success and failure $p_1$, $q_1$ and $p_2$, $q_2$, respectively, may be considered one compound experiment with four possible outcomes in each trial, namely, the combinations of (S,S), (S,F), (F,S) and (F,F). The assumption that the two original sequences are independent is translated into the statement that the probabilities of the four outcomes are $p_1$$p_2$, $p_1$$q_2$, $q_1$$p_2$, $q_1$$q_2$, ...
My thought was that I needed to calculate $p = P((S,F)|((S,F) or (F,S))) = \frac{p_1q_2(p_1q_2 + q_1p_2)}{(p_1q_2 + q_1p_2)} = p_1q_2$ however the book quotes the answer as $\frac{p_1q_2}{(p_1q_2 + q_1p_2)}$ but I can't quite understand why this is the case as the trials are independent.
Would anybody be able to help explain why this is the case? And also why any previous combinations affects the current probability at all? Thanks
Your formula for conditional probability is incorrect, and I'm not quite sure how you achieved it since your comment is a bit hard to read. Regardless it should be: $$\mathbb{P}(A | A \cup B) = \frac{\mathbb{P}(A \cap [A \cup B])}{\mathbb{P}(A \cup B)} = \frac{\mathbb{P}(A)}{\mathbb{P}(A \cup B)}$$ You can then do the same steps to achieve the correct answer. I think you made a mistake in the last step, where you assume that $A \cap (A \cup B)$ are two independent events, hence they can be multiplied. But this isn't true.
If $A$ is true, $A \cup B$ is also clearly true, so they must not be independent (the first event causes the second). However, lucky for us this is an identity called the absorption identity, and it's proof is straightforward and visible on this thread and it states that: $$A \cap (A \cup B) = A \cup (A \cap B) = A$$