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4 individuals arrive separately at an orchestra concert with

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4 individuals arrive separately at an orchestra concert with [#permalink] New post   08 Jun 2014, 10:42
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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



This problem came in veritas prep and i was stumped by this problem.
Would appreciate if someone is able to show how to solve this problem.

Kudos me if you like the question !!!!
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C
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Probability and Combinatorics (700-800 Level) [#permalink] New post   26 Jun 2014, 06:37
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francoimps 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

*Source Veritas Free Online Question Bank

I'm trying to use complementary probability but I can't seem to get the correct answer.

OA
Show Spoiler
C


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
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Re: 4 individuals arrive separately at an orchestra concert with [#permalink] New post   29 Jun 2014, 20:10
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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).
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Probability and Combinatorics (700-800 Level) [#permalink] New post   26 Jun 2014, 05:48
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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

*Source Veritas Free Online Question Bank

I'm trying to use complementary probability but I can't seem to get the correct answer.

OA
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C
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Re: 4 individuals arrive separately at an orchestra concert with [#permalink] New post   26 Jun 2014, 08:02
@desaichinmay22, great answer..

I was trying hard, but the point I missed was if A by chance sat in his own place, then B and C will also sit in their own places, thus leaving D to sit in his own place.. the randomness comes only when A does not sit in his place.
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Re: 4 individuals arrive separately at an orchestra concert with [#permalink] New post   26 Jun 2014, 21:24
Can one of the moderators explain this in more detail?
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Re: 4 individuals arrive separately at an orchestra concert with [#permalink] New post   29 Jun 2014, 16:33
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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).
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Re: 4 individuals arrive separately at an orchestra concert with [#permalink] New post   01 Jul 2014, 02:46
desaichinmay22 wrote:
francoimps 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

*Source Veritas Free Online Question Bank

I'm trying to use complementary probability but I can't seem to get the correct answer.

OA
Show Spoiler
C


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


Not clear on a few of your assumptions highlighted above. If a sits on correct seat , b and c could sit on each other's seat i.e. incorrect seat. However then d will sit on correct seat. Alternatively b and c could also sit on correct seats thereby all four sitting on correct seats.

So for basically d to be sitting in correct seat below are the combinations possible
a) all four correct seats
b) two correct and two incorrect seats.
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4 individuals arrive separately at an orchestra concert with [#permalink] New post   31 Aug 2014, 19:38
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desaichinmay22 wrote:
francoimps 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

*Source Veritas Free Online Question Bank

I'm trying to use complementary probability but I can't seem to get the correct answer.

OA
Show Spoiler
C


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


Can someone explain why the probability of A not selecting his assigned seat will be 3/4 when we want to ensure D gets to sit on his assigned seat? I am thinking 3/4 should be when A can select any of the 3 seats, besides the one he is assigned to. This includes the possibility that A CAN select D's seat as well. However, we are looking for cases where D gets to sit on his assigned seat so shouldn't second scenario be when A selects either B or C's seat (2/4); AND B selects either of the 2 remaining besides D (2/3); AND C selects the remaining seat besides D (1/2)

Which should lead to -> 2/4*2/3*1/2

Final answer 1/4 + 1/6 = 5/12?

where am I going wrong?

Thanks.
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4 individuals arrive separately at an orchestra concert with [#permalink] New post   28 Oct 2015, 05:34
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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
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Re: 4 individuals arrive separately at an orchestra concert with [#permalink] New post   28 Oct 2015, 23:06
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akhil911 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



This problem came in veritas prep and i was stumped by this problem.
Would appreciate if someone is able to show how to solve this problem.

Kudos me if you like the question !!!!


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)
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Re: 4 individuals arrive separately at an orchestra concert with [#permalink] New post   29 Oct 2015, 03:45
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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
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Re: 4 individuals arrive separately at an orchestra concert with [#permalink] New post   29 Oct 2015, 05:19
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Zhenek wrote:
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


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."
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4 individuals arrive separately at an orchestra concert with [#permalink] New post   29 Oct 2015, 05:24
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!?)
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Re: 4 individuals arrive separately at an orchestra concert with [#permalink] New post   29 Oct 2015, 05:50
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Zhenek wrote:
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!?)


Ok, so you are considering the sequence of writing the numbers as the sequence of seats. Yes, in that case, it certainly looks correct.
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Re: 4 individuals arrive separately at an orchestra concert with [#permalink] New post   07 May 2017, 22:06
1/2(1st person to sit correct or incorrect) * 1/2(2nd person to sit correct or incorrect) * 1/2(3rd person to sit correct or incorrect ) * 1 (Last person is definitely Correct )* 4( four Distinct people possible to come last) .

I don't think my approach is right.
Experts correct me if my logic is wrong.
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Re: 4 individuals arrive separately at an orchestra concert with [#permalink] New post   06 May 2020, 23:29
hubahuba wrote:
desaichinmay22 wrote:
francoimps 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

*Source Veritas Free Online Question Bank

I'm trying to use complementary probability but I can't seem to get the correct answer.

OA
Show Spoiler
C


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


Can someone explain why the probability of A not selecting his assigned seat will be 3/4 when we want to ensure D gets to sit on his assigned seat? I am thinking 3/4 should be when A can select any of the 3 seats, besides the one he is assigned to. This includes the possibility that A CAN select D's seat as well. However, we are looking for cases where D gets to sit on his assigned seat so shouldn't second scenario be when A selects either B or C's seat (2/4); AND B selects either of the 2 remaining besides D (2/3); AND C selects the remaining seat besides D (1/2)

Which should lead to -> 2/4*2/3*1/2

Final answer 1/4 + 1/6 = 5/12?

where am I going wrong?

Thanks.


you are missing one of the cases: (2/4)*(1/3)*(1/2)=1/12
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4 individuals arrive separately at an orchestra concert with [#permalink] New post   16 May 2020, 01:43
NUMBER OF WAYS TO ARRANGE THE 4 PEOPLE = 4*3*2*1 = 24
WHEN the LAST ONE SITS ON THE CORRECT SEAT, NO OF ARRANGEMENTS FOR THE FIRST THREE = 3*2*1 = 6
HENCE THE PROBABILITY THAT the LAST ONE WILL GET THE CORRECT SEAT = 6/24 = 1/4

CAN ANYONE DETECT THE MISTAKE?
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Re: 4 individuals arrive separately at an orchestra concert with [#permalink] New post   17 May 2020, 03:29
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Lokeshj039 wrote:
NUMBER OF WAYS TO ARRANGE THE 4 PEOPLE = 4*3*2*1 = 24
WHEN the LAST ONE SITS ON THE CORRECT SEAT, NO OF ARRANGEMENTS FOR THE FIRST THREE = 3*2*1 = 6
HENCE THE PROBABILITY THAT the LAST ONE WILL GET THE CORRECT SEAT = 6/24 = 1/4

CAN ANYONE DETECT THE MISTAKE?


Hi Lokesh,

You need to re-consider both the number of possible outcomes and number of expected outcomes. This problem statement has constraints defined, which you need to take into consideration when calculating the individual possibilities.

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Re: 4 individuals arrive separately at an orchestra concert with [#permalink] New post   10 Aug 2022, 11:13
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Lokeshj039 wrote:
NUMBER OF WAYS TO ARRANGE THE 4 PEOPLE = 4*3*2*1 = 24
WHEN the LAST ONE SITS ON THE CORRECT SEAT, NO OF ARRANGEMENTS FOR THE FIRST THREE = 3*2*1 = 6
HENCE THE PROBABILITY THAT the LAST ONE WILL GET THE CORRECT SEAT = 6/24 = 1/4

CAN ANYONE DETECT THE MISTAKE?


Hi Lokesh

we need to multiply 1/4 with 2! for the number of cases considering P4 got S1.
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Re: 4 individuals arrive separately at an orchestra concert with [#permalink]
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