$\begin{array}{1 1}(A)\;\large\frac{4}{5}\\(B)\;\large\frac{9}{5}\\(C)\;\large\frac{2}{3}\\(D)\;\large\frac{2}{5}\end{array} $

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- Required probability =$\large\frac{n(E)}{n(S)}$

Given six employees

Two are married to each other.

Tital number of outcomes they can sit randomly in a row n(S)=6!

Now probability of non adjacent desks =1-probability of adjacent desks

$\therefore$ Let us find the probability of adjacent desks.

Step 2:

There are two couples

$\therefore$ Number of outcomes of random arrangement of employees =5!

Consider 1 couple as 1 unit and arrangement of couple =2!

(ie) if a couple is AB then the seating arrangement can be AB or BA.

$\therefore$ Number of favorable outcomes n(E)=$5!\times 2!$

Step 3:

Required probability =$\large\frac{n(E)}{n(S)}$

$\Rightarrow \large\frac{5!\times 2!}{6!}$

$\Rightarrow \large\frac{2\times 1}{6}$

$\Rightarrow \large\frac{ 1}{3}$

Step 4:

$\therefore$ Probability of non-adjacent desks =1-probability of adjacent desks

$\Rightarrow 1-\large\frac{1}{3}$

$\Rightarrow \large\frac{2}{3}$

Hence (C) is the correct answer.

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