Question 44

The lattice $L = \{p, q, r, s, t\}$ is the standard diamond lattice $M_3$ with bottom $p$, top $t$, and three pairwise incomparable intermediate elements $q, r, s$.

Total number of triplets in $L^3$:

Testing the distributive identity $x \lor (y \land z) = (x \lor y) \land (x \lor z)$:

• If any of $x, y, z$ is $p$ or $t$, or if any two elements are equal, or if the elements are comparable, distributivity holds identically.
• In $M_3$, the distributive law fails only when $\{x, y, z\} = \{q, r, s\}$:
  For example, if $x = q, y = r, z = s$:
  $$x \lor (y \land z) = q \lor (r \land s) = q \lor p = q$$
  $$(x \lor y) \land (x \lor z) = (q \lor r) \land (q \lor s) = t \land t = t.$$
  Since $q \ne t$, distributivity fails for this triplet.
• Since there are $3! = 6$ permutations of distinct elements from $\{q, r, s\}$, distributivity fails for only a very small number of triplets (specifically 6 or 12 depending on operand order).
• Hence, the vast majority of triplets ($> 110$ out of $125$) satisfy the distributive equality.

The probability $p_r$ satisfies:

Since $\frac{1}{5} = 0.20 < 0.952 < 1$, we have:

This matches option D.