Suppose $\mathcal{L} = \{p, q, r, s, t\}$ is a lattice represented by the following Hasse diagram: For any $x, y \in \mathcal{L}$, not necessarily distinct, $x \vee y$ and $x \wedge y$ are join and meet of $x, y$, respectively. Let $\mathcal{L}^3 = \{(x, y, z) : x, y, z \in \mathcal{L}\}$ be the set of all ordered triplets of the elements of $\mathcal{L}$. Let $p_r$ be the probability that an element $(x, y, z) \in \mathcal{L}^3$ chosen equiprobably satisfies $x \vee (y \wedge z) = (x \vee y) \wedge (x \vee z)$. Then

The given Hasse diagram represents the classic non-distributive modular lattice $M_3$, where $p$ is the bottom element ($0$), $t$ is the top element ($1$), and $q, r, s$ are pairwise incomparable elements.

1. Total Number of Triplets:

   $$|\mathcal{L}^3| = 5^3 = 125$$

2. Check Distributivity ($p_r \neq 1$): For the triplet $(x, y, z) = (q, r, s)$:

   $$\text{LHS} = q \vee (r \wedge s) = q \vee p = q$$

   $$\text{RHS} = (q \vee r) \wedge (q \vee s) = t \wedge t = t$$

   Since $\text{LHS} \neq \text{RHS}$, the distributive identity is not satisfied for all triplets, so $p_r < 1$.

3. Satisfying Triplets ($p_r > \frac{1}{5}$):

   • Whenever $x = t$ (top element):

     $$t \vee (y \wedge z) = t \quad \text{and} \quad (t \vee y) \wedge (t \vee z) = t \wedge t = t$$

     This holds for all $5 \times 5 = 25$ pairs of $(y, z)$.

   • Whenever $x = p$ (bottom element):

     $$p \vee (y \wedge z) = y \wedge z \quad \text{and} \quad (p \vee y) \wedge (p \vee z) = y \wedge z$$

     This also holds for all $5 \times 5 = 25$ pairs of $(y, z)$.

   Since there are already at least $25 + 25 = 50$ distinct triplets satisfying the condition (and others where $y = z$, $x = y$, etc.), the number of satisfying triplets is strictly greater than $25$.

   Therefore:

   $$p_r > \frac{25}{125} = \frac{1}{5}$$

Hence, $\frac{1}{5} < p_r < 1$, which corresponds to option (D).