Question: $X$ speaks truth in $60\%$ and $Y$ in $50\%$ of the cases. What is the probability that they contradict each other narrating the same incident?
In this question , the textbook solution is that $$ \begin{split} p &= \mathbb{P}[\text{X and Y contradict each other}] \\ &= \mathbb{P}[\text{X lies and Y speaks truth}] + \mathbb{P}[\text{Y lies and X speak truth}] \\ &= \frac{6}{10} \times \frac{5}{10} + \frac{4}{10} \times \frac{5}{10} \\ &= \frac{1}{2}. \end{split} $$
However according to me should not the case of both lying together should also be added, as they both can contradict each other when both are lying at the same time, as well
The book is thinking of a simple problem with binary answers. For example, consider the following
Assume that both Mr. X and Mr. Y are asked about this question. The probability they will contradict each other will be exactly as the book is computing.
On the other hand, you are thinking of a much more vague situation, like
In such circumstances, Mr. X and Mr. Y can both lie and contradict each other, i.e., one says "cheese" and the other says "bread" when the actual breakfast was steak and potatoes.
In this situation, you are correct, but if you allow this much freedom, they can tell the truth and contradict each other as well, e.g., one will say "steak" and the other will say "potatoes." For this reason, to avoid any ambiguity, the book considers a much simpler setup.
In either case, I agree that the problem needs to be formulated more carefully to avoid the ambiguity in question.