There is a concept in Bayesian network literature, called I-equivalence. Two Bayesian network structures are called I-equivalence if they encode the same set of conditional independencies. For example, the following three structures are I-equivalence, since they all encode $A$ is independent of $C$ given $B$:
$$A\rightarrow B \rightarrow C$$
$$A\leftarrow B \leftarrow C$$
$$A\leftarrow B\rightarrow C$$
But the following structure does not belong to the I-equivalence class of the above three structures:
$$A\rightarrow B \leftarrow C$$
This is because of the v-structure or the head-to-head node $B$. In the above structure, $A$ is NOT independent of $C$ given $B$. There is a very useful theorem for checking the I-equivalency of two structures:
Two Bayesian network structures are I-equivalence if and only if they have the same set of immoralities and the same skeleton. Immoralities are head-to-head nodes without any edge between the parents. For example, $A\rightarrow B \leftarrow C$ is an immorality but it is not an immorality if there is an edge between $A$ and $C$. The skeleton of a Bayesian network structure is simply its undirected version.
Obviously, the I-equivalence relation is an equivalence relation which partition the space of structures into equivalence classes. In the above examples, $A\rightarrow B \leftarrow C$ belongs to another class than the class of other three structures. In your example, $Age \rightarrow Edu$ and $Age \leftarrow Edu$ belong to the same equivalence class.
No Bayesian network structure learning algorithm can choose a structure from the equivalence class based on data alone. In other words, the structures in an equivalence class cannot be distinguished based on data alone. Therefore, a Bayesian network structure learning cannot favor $Age \rightarrow Edu$ over $Age \leftarrow Edu$ or vice versa based on the data alone.
Note that this does not mean that the structure learning algorithms cannot find any direction in the graph. For example, if according to the data it finds out that the skeleton of the graph is $A-B-C$ and also $A$ is not independent of $C$ given $B$, it will conclude that there should be a v-structure, that is the correct structure is $A\rightarrow B \leftarrow C$.
Although it is not possible to choose a structure in an equivalence class based on data alone, as Diego mentioned we can exploit other knowledge than data to find the direction of undirected edges. For example, in our recent work , we tried to use the experts' knowledge to find more accurate Bayesian network structures.
Hope this short summary answers your question. For more information, I encourage you to read chapter 3 of the excellent book by Koller and Friedman .
 Amirkhani, Hossein, et al. "Exploiting Experts' Knowledge for Structure Learning of Bayesian Networks." IEEE Transactions on Pattern Analysis and Machine Intelligence (2016).
 Koller, Daphne, and Nir Friedman. Probabilistic graphical models: principles and techniques. MIT press, 2009.