How likely is it for Selmer groups to have mu invariant 0? Given a number field $K$, how likely is it that we'll find at least one elliptic curve $E/K$ such that the $\mu$-invariant of its Selmer group is 0 (in a cyclotomic extension)?
 A: Let's suppose you did not fix $p$, but you fixed $K$. I think that it is easier to find an elliptic curve with $\mu=0$ than to find an elliptic curve with rank zero and finite Tate-Shafarevich group. Indeed, if $E$ has rank zero and Sha is finite, then there is a prime $p$ such that


*

*$E$ has good ordinary non-anomalous reduction at all primes above $p$

*The $p$-primary part of Sha is trivial

*No Tamagawa number is divisible by $p$
(The lower two are ok and I would believe there is an argument for the top one, too. But I have not tought about it.)
It then follows that the $p$-primary Selmer group for that $E$ over the cyclotomic $\mathbb{Z}_p$-extension is finitely generated and torsion. The conditions imply that the leading term (=constant term) of the charactersitic series is a unit and hence the Selmer group is finite - and that is much more than just $\mu=0$.
If you wish to fix $p$, then you need to find a rank $0$ curve for which the above list of conditions holds. In any case, none of this can be done unconditionally because of the required finiteness of Sha.
I cannot see any easy way to show that $\mu=0$ other than showing that the Selmer group is finite alltogether. Conjecturally, for any given elliptic curve $E/K$ of any rank, the leading term of the $p$-adic charactersitic series is a unit for the vast majority of primes $p$ of good ordinary reduction (because of how we expect $p$-adic heights to be distributed). That still implies $\mu=0$. Therefore the answer to the question in the title is very very likely.
