= Solution
{c}
The matrices $X,Z$ are real symmetric, $\operatorname{tr}(X^2)=\operatorname{tr}(Z^2)=2$, and $\operatorname{tr}(XZ)=0$. Part <2/b> therefore gives
$$
\boxed{\langle\Omega|A\otimes B|\Omega\rangle
=\cos\alpha\cos\beta+\sin\alpha\sin\beta=\cos(\alpha-\beta).}
$$
Since the local measurement outcomes are $\pm1$, their product is one exactly on agreement. Thus $E[ab]=2\mathbb P(a=b)-1$, and
$$
\boxed{\mathbb P(a=b)=\frac{1+\cos(\alpha-\beta)}2
=\cos^2\!\left(\frac{\alpha-\beta}{2}\right).}
$$
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