From a group of M employees, N will be selected, at random,

Statement 1: insufficient since we need to know M to solve for the probability
Statement 2: N= M
This turns to sit N people into N chairs
For A is seated somewhere to the right of employee Georgia, we have (1+ 2+ 3+....+ (n-1)) possibilities
To illustrate this, lets say 4 people sit in 4 chairs
G in seat 1, A can be seated in 2,3, and 4 ( 3 possibilities)
G in seat 2, A can be seated in 3,4 ( 2 possibilities)
G in seat 3, A can be seated in 4 only (1 possibility)
Sum = 1+ 2+ 3
After G and A are seated, there are (n-2)! possible combination
( for example, in this case, after G and A are seated, there are two seats left, we have 2! possible combination)
The total possible combinations for N people in N chairs is N!
The possible combinations that meet the requirement is (1+2+3+...(N-1))* (N-2)! = [(N-1)*N/2] * (N-2)!
The probability = [(N-1)*N/2] * (N-2)!/N! = [(N-1)*N/2]/ [(N-1)*N] = 1/2
B is the answer.