= Solution
The unique maximum-$M$ product state is
$$
\boxed{|J,J\rangle=|j_1,j_1\rangle_1|j_2,j_2\rangle_2,
\quad J=j_1+j_2.}
$$
Apply the total lowering operator $J_-=J_{1-}+J_{2-}$ and normalize to obtain that multiplet. At the next highest remaining $M$, choose a normalized vector orthogonal to all previously constructed multiplets and annihilated by $J_+$. It is the next highest-weight state; lower it in the same way. Iterating produces every allowed total <angular momentum> and the <Clebsch-Gordan coefficients>.
For equal constituent spins, a total-spin singlet has $M=0$, so only the product basis vectors with opposite magnetic quantum numbers can occur. Acting with $J_+$ on the expansion and examining the coefficient of $|j,m+1\rangle_1|j,-m\rangle_2$ gives
$$
(\alpha_m+\alpha_{m+1})\sqrt{(j-m)(j+m+1)}=0,
\qquad -j\leq m<j.
$$
Thus $\alpha_{m+1}=-\alpha_m$. Normalization requires $(2j+1)|\alpha_m|^2=1$. Choosing a conventional overall phase gives
$$
\boxed{|0,0\rangle=\frac1{\sqrt{2j+1}}
\sum_{m=-j}^j(-1)^{j-m}|j,m\rangle_1|j,-m\rangle_2.}
$$
The exponent is an integer even for half-integral $j$. Projecting the specified extreme product state onto this unique singlet gives the measurement probability
$$
\boxed{\mathbb P(J=0)=\frac1{2j+1}.}
$$
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