Probability and Combinatorics

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Probability and Combinatorics

by francoimps » Thu Jun 26, 2014 4:44 am
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

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

OA C
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by GMATinsight » Thu Jun 26, 2014 5:13 am
EDITED!!!
Last edited by GMATinsight on Mon Jun 30, 2014 1:57 am, edited 5 times in total.
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by GMATinsight » Thu Jun 26, 2014 5:33 am
by mistake here is the answer of question "what's the probability of atleast one of them sitting at the right place if they randomly take places one after the other"


If A sits at right place then remaining 3 can sit in only 2 wrong ways
So last person taking right place and other three taking wrong seat will have = 4 x 2 = 8 ways

Now If A sits at right place and one other person also sits at right place then there is only one way of the remaining 2 sitting at wrong places so total such ways is equal to total ways of choosing 2 out 4 people who occupy right places = 4C2 = 6 ways

If all sit at right places then there is only one such possible way = 1 way

Total such cases = 8+6+1 = 15

Probability = 15/24 = 5/8 This should be the right answer
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by GMATGuruNY » Sat Jun 28, 2014 7:29 am
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
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.
Last edited by GMATGuruNY on Sun Jun 29, 2014 7:14 am, edited 1 time in total.
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by GMATGuruNY » Sun Jun 29, 2014 5:38 am
GMATinsight wrote:If D goes last then Probability of D occupying the last place will be 1/2 only as out of 24 outcomes noted here, in 12 outcomes D takes the right place and in remaining 12 D takes the wrong place

Therefore probability = 1/2
Can you expand upon your reasoning?
If there are no constraints on how people are seated, here are the 24 possible arrangements:

ABCD
ABDC
ACBD
ACDB
ADBC
ADCB

BCAD
BCDA
BDAC
BDCA
BACD
BADC

CDAB
CDBA
CBAD
CBDA
CADB
CABD

DABC
DACB
DCBA
DCAB
DBCA
DBAC

As indicated by the arrangements in red, D gets the correct seat in 6 -- not 12 -- of these 24 arrangements.
If there are no constraints on how people are seated, the probability of getting the correct seat is the same for each person:
1/4.
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by GMATinsight » Sun Jun 29, 2014 6:43 am
Hey Mitch...


The explanations is "If A, B and C sit randomly on 3 out of 4 places then probability of D's place to remain vacant is equal to probability of D's place to be occupied which is 1/2 and now D has only one place left to occupy so whether he want's or not he will have to occupy the only vacant place either right or wrong therefore the probability of D to occupy it's right place is same as the probability of D to occupy the wrong place which is 1/2"

Let me know if you find anything unconvincing...
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by GMATinsight » Sun Jun 29, 2014 6:58 am
Another approach is

Let 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) x (2/3) x (1/2) = 1/4

Total probability = (1/4)+(1/4) = 1/2
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by GMATGuruNY » Sun Jun 29, 2014 7:02 am
GMATinsight wrote:The explanations is "If A, B and C sit randomly on 3 out of 4 places then probability of D's place to remain vacant is equal to probability of D's place to be occupied which is 1/2"
This line of reasoning seems invalid.
If A, B and C sit randomly, the probability that D's seat remains vacant = 1/4, while the probability that D's seat is occupied = 3/4.
The two probabilities are not both equal to 1/2.
Please refer to my second post, which illustrates that D's seat remains vacant in only 6 of the 24 possible arrangements.

Regardless, B and C are not allowed to sit randomly, since each must take his assigned seat if it is available.
This constraint reduces the number of possible arrangements to 8, as shown in my initial post.
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by GMATinsight » Sun Jun 29, 2014 7:31 am
These are the 8 cases out of which, in half of the cases D's place is vacant and in the remaining half it's occupied by someone else.

ABC_
BAC_
CAB_
_ABC
_ACB
CBA_
_BAC
_BCA

But yes.. The line of reasoning was based on some mentally though of arrangements which could have led to some mistake as well.
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by GMATGuruNY » Sun Jun 29, 2014 10:31 am
GMATinsight wrote:These are the 8 cases out of which, in half of the cases D's place is vacant and in the remaining half it's occupied by someone else.

ABC_
BAC_
CAB_
_ABC
_ACB
CBA_
_BAC
_BCA

But yes.. The line of reasoning was based on some mentally though of arrangements which could have led to some mistake as well.
Isn't this solution quite different?
In the initial line of reasoning, A, B and C are randomly assigned, there are 24 possible arrangements, and it is suggested that D will be correctly placed in 12 of these arrangements.
In reality, if A, B and C are randomly assigned, then D will be correctly placed in only 6 of the 24 possible arrangements, leading to the selection of an incorrect answer choice (A):
6/24 = 1/4.
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by GMATGuruNY » Mon Jun 30, 2014 4:32 am
GMATinsight wrote: 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) x (2/3) x (1/2) = 1/4
Can you clarify the reasoning behind the fractions?
You seem to be calculating the probability that D gets his assigned seat if A takes the wrong seat.

Red fraction:
For D's seat to remain unoccupied, A must occupy either B's seat or C's seat.
Since A has only 2 options, shouldn't the first fraction be 2/4?

Blue fraction:
If A takes C's seat, then the probability that B takes A's seat or C's seat is 0, since B must take his own assigned seat if it is available.
The fraction in blue does not seem to account for this case.

Green fraction:
If A and B swap seats, then the probability that C takes D's seat is 0, since C must his own assigned seat if it is available.
The fraction in green does not seem to account for this case.
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