GATE CSE 2015 Set 1 | Question: 34

Question

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

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

Comments

@Kabir5454 Yes it was a typo. Sorry.. I have corrected it.

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if the same question is asked for pentagon structure what would be the probability?

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good expanation 

https://gateoverflow.in/8281/gate-cse-2015-set-1-question-34?show=107088#a107088

Answer 1

here , short cut to this question

by watching figure u can guess that it's not distributive lattice, becoz for q there is 2 complement so , it cant satisfy distributive property , therefore, probability cant be 1 , so option (a) is wrong

now if u apply distributive property for any 3 elements which are in straight line they will satisfy the distributive property , it concludes that its probability cant be 0, therefore option (b) is wrong

now no of possible distributive equation will be 5*5*5=125 because repetition is allowed here. since in above line, i hav proved that elements in straight line will follow the distributive prop then no of possible distributive with 3 element is 3*3*3=27 (since repeatition is allowed , u might ask question that there are 5 elements then why 3 ? since we have to fix 2 as universal lower bound and upper bound ) so 27/125 is greater than 1/5

 

so d option is correct

Answer 2

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