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

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

  1. $p_r = 0$
  2. $p_r = 1$
  3. $0 < p_r ≤ \frac{1}{5}$
  4. $\frac{1}{5} < p_r < 1$
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6 Answers

Best answer
94 votes
94 votes

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.
When $x = t,$ we have $5*5 = 25$ cases $($for any $y$ and $z$$)$ where the given conditions are satisfied. Now, with $x = r, y = p, z = p,$ we have one more case. So, $26/125$ which means strictly greater than $1/5.$
So, this makes $p_r > \frac{25}{125}$ 

Also,
for $x = q, y = r, z = s,$ the given condition is not satisfied as $q \vee (r \wedge s) = q \vee p = q,$ while $(q \vee r) \wedge  (q \vee s) = t \wedge t = t.$ So, $p_r \neq 1.$

These findings make option A, B, C as FALSE.

Hence, answer = option D

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125 votes
125 votes

Each element of {q,r,s} has two complement.So,these element can't be chosen in the triplets<x,y,z>.One element should be such that it has unique complement to meet the given distributive property.

So,ways can't be chosen = 3! = 6

(q,r,s)(q,s,r)(r,s,q)(r,q,s)(s,q,r)(s,r,q)

So,<x,y,z> be chosen 5$\times$5$\times$5=125 ways.

So,Ans = $\frac{125-6}{125}$ = $\frac{1}{5}\lt\frac{119}{125}\lt1$       (D)

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18 votes
18 votes

For p=1 we have to satisfy the condition of distributive lattice i.e. x V (y ⋀ z) = (x V y ) ⋀ (x V z) 

but it is not distributive as q has here more than one complement.

If we take (q,r,s) and apply distributive formula, we can see it is not satisfying i.e.  x V (y ⋀ z) ≠ (x V y ) ⋀ (x V z) 

So, from here we can say obviously pr < 1

Here(x,y,z) can take p,q,r,s,t i.e. 5 values in 5*5*5=125 ways

Now, for proving  pr >1/5 we have to check some triplets which satisfying distributive conditions, 

take for t 

(t,t,t),(t,t,q),(t,t,r),(t,t,s),(t,t,p),

(t,q,t),(t,q,q),(t,q,r),(t,q,s),(t,q,p),

(t,r,t)................................(t,r,p),

(t,s,t)................................(t,s,p),

(t,p,t).................................(t,p,p)

these 25 points satisfied distributive property, right?i.e. pr =25/125=1/5

Now take 1 more point to say pr>1/5

take any  satisfactory point .e.g  take (r,r,p) 

So, we can easily say 1/5 < p< 1

Ans (D)

17 votes
17 votes

“$x∨(y∧z)=(x∨y)∧(x∨z) $”

This is distributive property of lattices. If a lattice is a distributive lattice, then each element has at most one complement (ie, $0$ or $1$ complements)

In the given lattice, $p$ and $t$ have $1$ complement each. $q,$r and $s$ have two complements each.

In the triplet, even if one element has $0$ or $1$ complements, the distributive property is satisfied.

Hence, only triplet that isn't allowed is $(q,r,s)$

 

But hold on. The question says that the triplets are ordered. Hence, $(q,r,s)$ is different from $(r,s,q)$ and both won't work.

Total triplets that aren't favourable = $3!=6$

So, favourable triplets = $|total|-6$

 

$|total| \neq {^5C}_3$ because the elements are "not necessarily distinct".

$|total|=5*5*5=125$

 

So, favourable triplets = $125-6=119$

 

Probability of favourable triplets = $\frac{119}{125}$, which is greater than $\frac{1}{5}$ but smaller than $1$

 

So, Option D

Answer:

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