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Question

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

For any xyϵL, not necessarily distinct, xy and xy are join and meet of x,y respectively. Let L3={(x,y,z):x,y,zϵL} be the set of all ordered triplets of the elements of L. Let pr, be the probability that an element (x,y,z)ϵL3 chosen equiprobably satisfies x(yz))=(xy)(xz). Then

A
pr=O
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B
pr=1
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C
0<pr15
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D
15<pr<1
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Solution

The correct option is D 15<pr<1
L has 5 elements
L3 has all ordered triplets of elements of L
L3 contain 5×5×5=53=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!=3×2×1=6
p(satisfying distributive property)
=1p(violate the distributive property)
=1653=0.952 which is between 15 and 1.

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