= Solution
Apply the gates from left to right. The first <Hadamard gate> and the first <controlled-NOT gate> produce the <Bell state>
$$
\frac{|00\rangle+|11\rangle}{\sqrt2}.
$$
The parallel $Z\otimes H$ gates then produce
$$
\frac12\left(|00\rangle+|01\rangle-|10\rangle+|11\rangle\right).
$$
The final controlled-NOT has the lower qubit as control; it interchanges $|01\rangle$ and $|11\rangle$, whose amplitudes here are equal. Thus the output is
$$
\boxed{
|\psi_{\rm out}\rangle
=\frac12\left(|00\rangle+|01\rangle-|10\rangle+|11\rangle\right).}
$$
By <quantum measurement in the computational basis>, every basis outcome has probability equal to the squared modulus of its amplitude:
$$
\boxed{\Pr(00)=\Pr(01)=\Pr(10)=\Pr(11)=\frac14.}
$$
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