If you have a sample space $\Omega$, then you can consider an event to be either a subset of $\Omega$, or a logical formula $\phi(x)$ parameterized by an element $x$ of $\Omega$ (here we might formally consider $x$ an $\Omega$-valued random variable, namely the identity $\Omega\to\Omega$).
These two views are equivalent; for $A\subseteq\Omega$ we can write either $P(A)$ or $P(x\in A)$; and for $\phi(x)$ we can write either $P(\phi(x))$ or $P(\{x\in\Omega\mid\phi(x)\})$. So it doesn't really matter which convention one uses, since it is easy to convert from one or the other.
The core definitions in textbooks will usually use the subset-of-$\Omega$ formalism, because we're familiar with using sets as mathematical objects everywhere in mathematics, whereas using a logical formula as a mathematical object that can be manipulated is somewhat of a specialty. Also, there are technical reasons (measure theory) why one might not want all subsets of $\Omega$ to be events, and such restrictions are much easier to express in a set-theoretic language than as restrictions on which logical formulas are allowed. On the other hand, it takes quite a bit of devious ingenuity to build a purely logical $\phi(x)$ that does not represent an event, so this is less of a problem in practice than it is in theory.
As you have found out, these reasons often do not stop textbook authors from using logical notation when it suits them, and expect readers to be able to translate into the set-based formalism.
As additional abuse of notation, it is not uncommon to switch between the two viewpoints in the middle of a formula, such as in $P(\neg A\land \neg B)$, where $A$ and $B$ might have been defined as sets, but are used in a context where one must have a logical formula. In this case, too, you're suspected to mentally insert the appropriate conversion, giving $P(\neg(x\in A)\land \neg(x \in B))$.
First lets find the probability that you win on your $k^{th}$ draw. Let us denote it by $p_k$. Then the probability you win is given as $\displaystyle \sum_{k=1}^{\infty} p_k$.
The probability you win on your $k^{th}$ draw can be computed as follows. If you have to win on your $k^{th}$ draw, then in each of your previous $k-1$ draws, you should have chosen one of the balls marked LOSE else you would have won before your $k^{th}$ draw. Your opponent must have also chosen one of the balls marked LOSE in each of his $k-1$ draws else he would have won before you. In each draw, the probability that a person chooses one of the balls marked LOSE is $\dfrac45$ and the probability that a person chooses the ball marked WIN is $\dfrac15$. Hence,
\begin{align}
p_k & = \underbrace{\left(\dfrac45 \times \dfrac45 \times \cdots \times \dfrac45 \right)}_{\substack{(k-1) \text{ times}\\ \text{You lost your first $k-1$ draws}}}
\times \underbrace{\left(\dfrac45 \times \dfrac45 \times \cdots \times \dfrac45 \right)}_{\substack{(k-1) \text{ times}\\ \text{Opponent lost his first $k-1$ draws}}} \times \underbrace{\dfrac15}_{\text{You chose the right ball on your $k^{th}$ draw}}\\
& = \left(\dfrac45 \right)^{2k-2} \times \dfrac15
\end{align}
Hence, the probability you win is $$\sum_{k=1}^{\infty} p_k = \sum_{k=1}^{\infty} \left(\dfrac45 \right)^{2k-2} \times \dfrac15 = \dfrac15 \left( \sum_{k=1}^{\infty} \left(\dfrac{16}{25} \right)^{k-1}\right) = \dfrac15 \left( \dfrac1{1-\dfrac{16}{25}} \right) = \dfrac59$$
Best Answer
The complement event of both are not lawyers is not both are lawyers. The complement is at least one is lawyer.
Also, there is no reason to believe that $P(A \cap B)=0$ are disjoint.