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From a group of J employees, K will be selected, at random,

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From a group of J employees, K will be selected, at random,

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From a group of J employees, K will be selected, at random, to sit in a line of K chairs. There are absolutely no restrictions, either in the selection process nor in the order of seating - both are entirely random. What is the probability that the employee Lisa is seated exactly next to employee Phillip?

1) K = 15
2) K = J

The OA is C

Source: Magoosh

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From a group of J employees, K will be selected, at random, to sit in a line of K chairs. There are absolutely no restrictions, either in the selection process nor in the order of seating - both are entirely random. What is the probability that the employee Lisa is seated exactly next to employee Phillip?

We need both K, J values in order to find the probability
1) K = 15
Still J is not given
Insufficient

2) K = J
We can group Lisa and Phillip together and consider them as a single person, then we can arrange them in (K-1)!*2
But K value is unknown
Insufficient

1 and 2 together, (K-1)!*2
= 14!*2
Sufficient

Option C is correct

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Statement 1
K=15
From J; 15 people are randomly selected to seat into 15 chairs but we don't know the exact value of of J. If J is much larger than 15, then it is likely that Lisa and Philip are not among the 15 employees that were randomly selected to seat in 15 Chairs, the two of them would have to be selected before we find out if they are actually sitted exactly next to to each other without knowing the value of J.
Statement 1 is INSUFFICIENT.

Statement 2
K=J
If K=J then all employees are selected and seated
If K=J is a smaller number like 4 or 3 :it is very certain that Lisa and Philip will be more likely to seat next to each other but if (K=J) is a Larger number like 100, then it is more likely that Lisa and Philip are seated far away from each other, without knowing the value of either J or K : statement 2 is clearly INSUFFICIENT.

combining 2 statement together
k = 15 and K=J hence (K=J=15)
All 15 employees are seated
combining the two employees, we can arrange 14 people in 14 seats
$$\Pr obability\ that\ Lisa\ and\ Philip\ seat\ together\ =\frac{14!}{15!}$$ $$=\frac{14\cdot13\cdot12\cdot11\cdot10\cdot9\cdot8\cdot7\cdot6\cdot5\cdot4\cdot3\cdot2\cdot1}{15\cdot14\cdot13\cdot12\cdot11\cdot10\cdot9\cdot8\cdot7\cdot6\cdot5\cdot4\cdot3\cdot2\cdot1}$$
$$\frac{87178291200}{1307674368000}=\frac{1}{15}$$
$$\Pr obability\ that\ Lisa\ and\ Philip\ seat\ together=1-\frac{1}{15}=\frac{14}{15}$$
$$both\ statements\ together\ is\ SUFFICIENT\ $$
$$answer\ is\ Option\ C$$

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