For p_{r }=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 p_{r} < 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 p_{r} >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. p_{r} =25/125=1/5
Now take 1 more point to say p_{r}>1/5
take any satisfactory point .e.g take (r,r,p)
So, we can easily say 1/5 < p_{r }< 1
Ans (D)