Question

Suppose $L=\{p,q,r,s,t\}$ is a lattice represented by the following Hasse diagram:

For any $xyϵL$, not necessarily distinct, $x\vee y$ and $x\wedge y$ are join and meet of $x,y$ respectively. Let $L^3=\left\{\right(x,y,z):x,y,zϵL\}$ be the set of all ordered triplets of the elements of $L$. Let $p_r$, be the probability that an element $(x,y,z)ϵL^3$ chosen equiprobably satisfies $x\vee (y\wedge z))=(x\vee y)\wedge (x\vee z)$. Then

• A. $p_r=O$
• B. $p_r=1$
• C. $0<p_r\le \frac{1}{5}$
• D. $\frac{1}{5}<p_r<1$

Answer 1

The correct option is D $\frac{1}{5}<p_r<1$

$L$ has 5 elements
$L^3$ has all ordered triplets of elements of $L$
$\Rightarrow L^3$ contain $5\times 5\times 5=5^3=125$ elements.

If $q,r,s$ are chosen, then only it will violate the distributive property.
Number of ways to choose $q,r,s$ in any triplet order $=3!=3\times 2\times 1=6$
$\therefore$ p(satisfying distributive property)
$=1-p$(violate the distributive property)
$=1-\frac{6}{5^3}=0.952$ which is between $\frac{1}{5}$ and $1$.