Problem Solving

4 individuals arrive separately at an orchestra concert with assigned seating tickets for exactly 4 seats in a special section. The first person to arrive loses his ticket stub after entry but remembers the section and sits randomly in one of the 4 seats. After that, each person arrives and takes his or her assigned seat in the section if it is unoccupied, and one of the unoccupied seats at random otherwise. What is the probability that the last person to arrive gets to sit in his assigned seat?

A. 1/4
B. 3/8
C. 1/2
D. 5/8
E. 3/4

Ans: C

anksm22 wrote:4 individuals arrive separately at an orchestra concert with assigned seating tickets for exactly 4 seats in a special section. The first person to arrive loses his ticket stub after entry but remembers the section and sits randomly in one of the 4 seats. After that, each person arrives and takes his or her assigned seat in the section if it is unoccupied, and one of the unoccupied seats at random otherwise. What is the probability that the last person to arrive gets to sit in his assigned seat?

A. 1/4
B. 3/8
C. 1/2
D. 5/8
E. 3/4

Here, the number of ways to arrange the 4 people ≠ 4!.
The reason is that the 4 people CANNOT seat themselves randomly.
Once the first person is seated, every subsequent person MUST choose his assigned seat if it is available, REDUCING the total number of possible arrangements.

Let the 4 people, in order of arrival, be A, B, C, D.

Case 1: A takes the correct seat
A _ _ _
Since B's seat is available when he arrives, B takes the correct seat.
A B _ _
Since C's seat is available when he arrives, C takes the correct seat, yielding the following arrangement:
A B C D.

Case 2: A takes B's seat
_ A _ _

Case 2a: A takes B's seat, B takes A's seat
B A _ _
Since C's seat is available when he arrives, C takes the correct seat, yielding the following arrangement:
B A C D.

Case 2b: A takes B's seat, B takes C's seat
_ A B _
Since C's seat is NOT available when he arrives, C takes either A's seat or D's seat, yielding 2 possible arrangements:
C A B D
D A B C.

Case 2c: A takes B's seat, B takes D's seat
_ A _ B
Since C's seat is available when he arrives, C takes the correct seat, yielding the following arrangement:
D A C B.

Case 3: A takes C's seat
_ _ A _
Since B's seat is available when he arrives, B takes the correct seat.
_ B A _
Since C's seat is NOT available when he arrives, C takes either A's seat or D's seat, yielding 2 possible arrangements:
C B A D
D B A C.

Case 4: A takes D's seat
_ _ _ A
Since B's seat is available when he arrives, B takes the correct seat.
_ B _ A
Since C's seat is available when he arrives, C takes the correct seat, yielding the following arrangement:
D B C A.

Possible arrangements:
A B C D
B A C D
C A B D
D A B C
D A C B
C B A D
D B A C.
D B C A.

D gets the correct seat in 4 of the 8 possible arrangements.
Thus:
P(D gets the correct seat) = 4/8 = 1/2.

The correct answer is C.