Suppose $L = \left\{ p, q, r, s, t\right\}$ is a lattice represented by the following Hasse diagram:
For any $x, y ∈ L$, not necessarily distinct , $x ∨ y$ and $x ∧ y$ are join and meet of $x, y$, respectively. Let $L^3 = \left\{\left(x, y, z\right): x, y, z ∈ L\right\}$ be the set of all ordered triplets of the elements of $L$. Let $p_{r}$ be the probability that an element $\left(x, y,z\right) ∈ L^3$ chosen equiprobably satisfies $x ∨ (y ∧ z) = (x ∨ y) ∧ (x ∨ z)$. Then
- $p_r = 0$
- $p_r = 1$
- $0 < p_r ≤ \frac{1}{5}$
- $\frac{1}{5} < p_r < 1$
My attempt:
Number of elements in $L^3 =$ Number of ways in which we can choose 3 elements from 5 with repetition $= 5 * 5 * 5 = 125.$
Now, when we take $x = t,$ then the given condition for $L$ is satisfied for any $y$ and $z.$ Here, $y$ and $z$ can be taken in $5 * 5 = 25$ ways.
Take $x = r, y = p, z = p.$ Here also, the given condition is satisfied. So, $pr > \frac{25}{125} > 1/5.$
For $x = q, y = r, z = s$, the given condition is not satisfied as $q ∨ (r ∧ s) = q ∨ p = q$, while $(q ∨ r) ∧ (q ∨ s) = t ∧ t = t.$ So, $pr ≠ 1.$
Hence , option $(4)$ is true.
Can you explain in formal way, please?