Why there is a probability in the completeness part?
The standard definition of $\mathsf{IP}$ involves probability both in the completeness and soundness requirements. Removing randomness all together reduces $\mathsf{IP}$ to $\mathsf{NP}$ (the witness is the full accepting transcript). 

Note that you can amplify the success probability by "repeating the protocol" (i.e. talking with a prover $P'$ who simulates repetitions) and taking the majority vote (you need to be slightly careful here, since your dealing with a special prover who has access to previous interactions, but you're saved by the universal quantifier in the soundness requirement).

If what's bothering you is the fact that the protocols you are familiar with satisfy completeness with probability $1$, then one example is the proof for graph non isomorphism being in $\mathsf{AM[2]}$, using the set lower bound protocol, where the completeness probability is smaller than 1.

That said, you can in fact remove the probability from the completeness requirement (i.e. say that there exists a prover which convinces $V$ with probability $1$, $V$ is still probabilistic) without changing the class $\mathsf{IP}$, but this is a non trivial fact, which follows from the protocol for TQBP in the $\mathsf{IP=PSPACE}$ proof (which succeeds with probability 1 in case the formula is true).