4 individuals arrive separately at an orchestra concert with

Question

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

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

desaichinmay22 · Accepted Answer

Let us assume that there are a,b,c and d individuals. a is the first one to arrive at concert and d is the last one to arrive at concert.
If a sits on his assigned seat, then b and c will also occupy their assigned seats and d will get to sit on his assigned seat.
Probability of a occupying his assigned seat=1/4

If a doesn't occupy his assigned seat, then b will have to occupy seat of either a or c and c will have to occupy the seat that is not assigned to d from the remaining two.
Probability of happening this event=3/4*2/3*1/2=1/4

Total probability=1/4+1/4=1/2

Ans=C

sesh · Answer

This difficult probability question requires that you consider each of 4 possible scenarios. For purposes of illustration let's assign people letters and seats numbers and say that A is assigned 1, B is assigned 2, C is assigned 3, and D is assigned 4. Let's break it down into 4 equal buckets of 25% each determined by which seat A picks.

A PICKS 1:

There is a 100% probability that D gets the right seat if Person A picks 1, so this represents overall a 25% favorable probability (1 of the 4 possibilities of the seat A could pick)

A PICKS 2:

If B picks 1, C picks 3, then D gets correct seat.

If B picks 3 and C picks 1, then D gets correct seat.

If B picks 3 and C picks 4, then D gets wrong seat

If B picks 4, C gets 3, then D gets wrong seat

Overall, 2/4 possibilities are favorable so take ½ of 25% (overall 12.5% favorable probability)

A PICKS 3:

B gets 2 and C picks 1, then D gets correct seat.

B gets 2 and C picks 4, then D gets wrong seat.

Overall, ½ possibilities are favorable so take ½ of 25% (12.5% favorable probability)

A PICKS 4:

100% probability that D gets wrong seat (0% favorable probability)

Add all of these up and you will see that the probability of D getting the correct seat is 50%. Correct answer is (C).

raghavsatwik · Answer

According to me, the probability of a, b, c or d occupying their assigned seats is 1/2. (because they can either occupy seats assigned to them, or one of the non-assigned seats). So, probability of D occupying assigned seat is 1/2 (for that matter, probability of any one occupying his or her assigned seats is 1/2).

jmikery · Answer

Seems like the scenarios are simple enough to do a decision tree:

Add the following scenarios where the last person gets the correct seat:

The probability that the first person took the correct seat: 1/4

The probability that the first person took the seat of the 3rd person to arrive, the 2nd person takes their own seat, and the 3rd person takes the seat of the first person: 1/4*1/1*1/2=1/8

The probability that the first person took the seat of the 2nd person, the 2nd person takes the seat of the first person, and the 3rd person takes his correct seat: 1/4*1/3*1/1=1/12

The probability that the first person took the seat of the 2nd person, the 2nd person takes the seat of the 3rd person, and the 3rd person takes the seat of the first person: 1/4*1/3*1/2=1/24

1/4+1/8+1/12+1/24=6/24+3/24+2/24+1/24=12/24 or 1/2

KarishmaB · Answer

You can also use the process of elimination in this question.

Say there are four people P1, P2, P3 and P4 depending on who arrives first, second etc. They have their assigned seats S1, S2, S3 and S4.

The probability of P1 occupying S1 is 1/4 - in this case P4 will get S4 (Probability 1/4)

The probability of P1 occupying S4 is 1/4 - in this case P4 will not get S4 (Probability 1/4)

The probability of P1 occupying S3 is 1/4 - in this case P3 will take either S1 or S4, both with probability 1/2.
So there is 1/8 probability that P4 will get S4 and 1/8 probability that P4 will not get S4.

Note that 1/4 + 1/8 = 3/8 and you have the case leftover where P1 takes S2.
So the probability of P4 getting S4 will be more than 3/8 and less than 5/8.

The only answer is 1/2

Answer (C)

Zhenek · Answer

I found it fairly easy to just go with the bruteforce method of calculating the cases. Logic seems to be very strict so the amount of cases wasn't even that big (8 total), and as it so happened only 4 of those 8 turned out to be "good", thus 1/2.
Say seats are labeled as 1 2 3 4 and so are people, here are cases
1 2 3 4
---------
1)1 2 3 4 -good
2)2 1 3 4 - good
2)3 1 2 4 - good
4)4 1 2 3 - bad
5)4 1 3 2 - bad
6)3 2 1 4 - good
7)4 2 1 3 - bad
8)4 2 3 1 - bad

4 "good" out of 8, 1/2

KarishmaB · Answer

You missed some cases:

2 3 4 1
or
2 3 1 4
or
2 4 3 1
etc

Also, 3 1 2 4 (your number 3 case above) is not a possible case. There may be others. I am not checking all.

Focus on - "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."

Zhenek · Answer

I thought that they arrived one after another in a set in stone sequence (we can sort of assume so), as in 1 - arrives first, 2 - arrives second, 3 - arrives third, 4 - arrives last (fourth). So a case of 2 3 4 1 can't happen coz if first guy arrives and takes 4th spot, the 2nd guy will take his spot (spot #2) coz its free.

Going by my logic for case 3 1 2 4: first guy takes 2nd spot, 2nd guy has to take random spot (he takes 3rd), 3rd guy has to take random spot (he takes first), 4th guy takes his spot (lucky him!?)

KarishmaB · Answer

Ok, so you are considering the sequence of writing the numbers as the sequence of seats. Yes, in that case, it certainly looks correct.

anish777 · Answer

this can be solved using the derangement process as well. Using derangement, the total number of ways in which 4 of them sit in the wrong seats is 15. All of them sitting in the right way is 1. Total number of ways is 16 then.
Now 3 of them can sit in the right way = 1.
Using derangement, 3 of them sit in the wrong way = 4.
Using derangement along with combination, 2 of them sit in the wrong way = 3
Total = 8
Probability = 8/16 = 1/2.

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