There are two machines, machine A and machine B whose lifetimes are exponentially distributed with the same ${lambda}$ value. Given that one machine stopped functioning, what is the probability that the remaining functional machine is A?
My attempt is as follows:
The desired probability is the same as the probability that A is functioning and B is not, which is equivalent to, for all ${0 < t < \infty}$,
$$ \begin{align*} \int_{0}^{\infty} P\{A > t; B \le t \} &= \int_{0}^{\infty} P\{ A > t\}P\{ B \le t\} \\&= \int_{0}^{\infty} e^{-\lambda t} ( 1 - e^{-\lambda t}) \\&=\frac{1}{2\lambda}. \end{align*} $$
But the answer is 0.5.
Am I missing something?
I will illustrate your error with an example from a discrete distribution.
Suppose instead you have two coins, A and B, but they are unfair: on any given flip, the probability of obtaining heads is $p = 0.1$. On each turn, you flip them simultaneously. The random number of turns it takes for you to get heads for coin A is a geometric random variable, $$\Pr[A = a] = (0.1)(0.9)^{a-1}, \quad a \in \{1, 2, \ldots \}.$$ And the same thing is true of B: $$\Pr[B = b] = (0.1)(0.9)^{b-1}, \quad b \in \{1, 2, \ldots \}.$$ You flip the pair of coins until you observe at least one of them is heads--this means A or B or both are heads on some turn, say $n$.
Now, given that you stopped on this first instance of observing heads, what is the probability that coin A is showing tails?
According to your reasoning, you would compute
$$\sum_{n=1}^\infty \Pr[A > n]\Pr[B \le n]. \tag{1}$$ It's easy to see that $$\Pr[A > n] = (0.9)^n, \\ \Pr[B \le n] = 1 - (0.9)^n.$$ But this gives us
$$\sum_{n=1}^\infty (0.9)^n (1 - (0.9)^n) = \frac{90}{19} > 1.$$ This is impossible: we are interested in a probability. And now you might have an idea as to why $(1)$ is incorrect: it is a sum of probabilities of events that are not mutually disjoint; e.g., the outcome $(A = 8) \cap (B = 3)$ is included in both $(A > 5) \cap (B \le 5)$ and $(A > 6) \cap (B \le 6)$, so the sum $(1)$ is counting the probability of this outcome more than once.
For the same reason, your integral $$\int_{t=0}^\infty \Pr[A > t]\Pr[B \le t] \, dt$$ does not represent a probability.
The correct expression arises from carefully considering the set of desired outcomes of the joint variables $(A, B)$. The sample space of elementary outcomes is the set $\{(A,B) \in \mathbb R : (A \ge 0) \cap (B \ge 0)\}$. The set of outcomes for which $A$ is still functioning after $B$ has stopped is characterized by the condition $A > B$. Thus, the desired probability is $$\Pr[A > B] = \int_{a = 0}^\infty \int_{b=0}^a f_{A,B}(a,b) \, db \, da, \tag{2}$$ where $$f_{A,B}(a,b) = f_A(a) f_B(b) = \lambda e^{- \lambda a} \lambda e^{-\lambda b} \tag{3}$$ is the joint density of $(A,B)$. The details of the computation are straightforward.
Of course, none of this is intended to be the simplest or most elegant solution: the symmetry between $A$ and $B$ is obvious and so is the desired probability. However, the equation $(3)$ does suggest a way to generalize the problem; e.g., suppose $A$ is exponentially distributed with rate $\lambda_A$ and $B$ is exponentially distributed with rate $\lambda_B$, not necessarily equal to $\lambda_A$. Then $(2)$ and $(3)$ give us an immediate way to calculate the desired probability whereas an argument by symmetry is not possible.