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There are $k(G)|G|$ commuting pairs $(y,z)$ of elements of $G.$ How many commuting triples $(x,y,z)$ of elements of $G$ are there? If we fix the first component $x$ of the triple, note that $x$ permutes the commuting pairs of elements of $G,$ and the only such pairs it fixes are the commuting pairs which already have both components in $C_{G}(x).$ Hence $x$ fixes $k(C_{G}(x))|C_{G}(x)|$ commuting pairs, and the total number of commuting triples in $G \times G \times G$ is $\sum_{x \in G} k(C_{G}(x))|C_{G}(x)|$. If $G$ has $k$ conjugacy classes, with representatives $\{ x_{i} : 1 \leq i \leq k \}$, this may be rewritten as $|G| \sum_{i=1}^{k} k(C_{G}(x_{i})).$ The probability you require, assuming a uniform distribution, is $\frac{1}{|G|^{2}}\left( \sum_{i=1}^{k} k(C_{G}(x_{i})) \right)$.