More details about the Baillie–PSW test
The advantage in using base 2 is that we know all of the psp's base 2 up to $2^{64}$. It has been verified that none of these psp(2)'s passes a Lucas test when the parameters $P, Q$ are chosen in accord with any of the methods in the Baillie/Wagstaff paper.

If you choose a random base, there might be some composite $n$ that passes both the Fermat and Lucas tests. For example, $n = 5777$ is a strong psp base 76, and is also a Lucas pseudoprime.

By the way, if you implement a Lucas test, I would also recommend adding the following check, which is virtually free once you reach the end of the Lucas calculation. If $n$ is an odd prime and $(n, QD) = 1$ where $D = P^2 - 4Q$, (and, as usual, $\left(\frac{D}{n}\right) = -1$ (a Jacobi symbol), then $V_{n+1} \equiv 2Q \pmod n$. If $D, P$, and $Q$ are chosen by method $A^*$ (see Baillie/Wagstaff), then 913 is the only odd composite number up to 25 billion for which this congruence holds. (The B/W paper gives a limit of $10^8$, but I've recently carried the calculation farther). So, in addition to the sprp(2) and slprp(P,Q) tests, this congruence adds additional strength to the primality test.