Conditional Probability not giving desired result So I am not getting the given solutions to this problem and I would appreciate some feedback on where the errors are.
On any given day, the probability that Tomas works on his poetry is $0.55$, and the probability that he stops at his local pub is $0.75$. Given that he does at least one of those two things, find the probability that on any given day Tomas:
(i) Goes to the pub and works on his poetry
(ii) Goes to the pub but does not work on his poetry
So first I defined both events. Keeping in mind that this is conditional probability as he is given to do at least one of these two events.
Let $X$ be the event Tomas goes to the pub and let $Y$ be the event Tomas works on his poetry. He always does at least one of these two which is $P(\text{At least 1}) = 1 - P(\text{None}) = 1 - 0.25\times0.45=0.8875$.
Question (i) is essentially asking for
\begin{align}
P(X \text{ AND } Y | \text{ At least 1 }) &= \frac{P(\text{ At least 1 } | X \text{ AND } Y)P(X \text{ AND } Y)}{P(\text{ At least 1)}}\\
 &= \frac{1\times(0.75\times0.55)}{0.8875}\\
 &= 0.46478 \ldots  
\end{align}
This does not match the provided solution of $0.3$
Similarly for Question (ii)
\begin{align}
P(X \text{ AND } Y' | \text{ At least 1 }) &= \frac{P(\text{ At least 1 } | X \text{ AND } Y')P(X \text{ AND } Y')}{P(\text{ At least 1)}}\\
 &= \frac{(0.75 \times 0.45)\times(0.75 \times 0.45)}{0.8875}\\
 &= 0.12834\ldots
\end{align}
Which does not match the given solution of $0.45$.
Is there a flaw in my thinking or are the provided solutions incorrect?
 A: Examining the comments, let me first say that I originally answered part (1) in the same way that the OP (i.e. original poster) did.
That is, I assumed that the events of going to the pub and working on poetry were independent events, which led me to the conclusion that the desired probability is
$$ \frac{(0.55) \times (0.75)}{1 - [(0.45)(0.25)]} \approx 0.4648. \tag1 $$
When I then read that the given solution to part (1) was (instead) $(0.3)$ the challenge of part (1) changed.  Now, I had to accept that the problem was poorly written and had to reverse engineer some way of interpreting the hash so as to reach the computation of $(0.3).$

What you have to do is to assume that Tomas always either writes poetry or goes to the pub or both, every day.  You then have to assume that simultaneously, his probability of writing poetry and going to the pub are
$(0.55), (0.75)$ respectively.
Let $E_1$ denote the event of writing poetry.
Let $E_2$ denote the event of going to the pub.
What you then have to conclude is that since 
$(1.00) = P(E_1 \cup E_2) = p(E_1) + p(E_2) - p(E_1 \cap E_2)$ 
$= (0.55) + (0.75) - p(E_1 \cap E_2)$,
that you must have that $p(E_1 \cap E_2) = (0.3).$

Then, part (2) becomes $p(E_2) - p(E_1 \cap E_2)$ 
$= (0.75 - 0.3) = (0.45).$

Not all people knowledgeable in Math should write Math books.
