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## Probability: getting a seat at an overcrowded table!!

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Cúchulainn Just gettin' started!
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Probability: getting a seat at an overcrowded table!! Tue May 15, 2012 3:47 am
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Hi - i have just joined this forum, so I hope i'm using it correctly. I've just done a problem on one of the kaplan quizzes, which made me think of my own slightly different version, and was hoping someone could confirm my approach to my own version is correct or not!

A square table seats 1 person on each of the four sides, with every person directly facing another person across the table. If there are eight people looking to sit at this table and choose their seats at random, what is the probability that any two of the 8 getting seats directly facing each other?

My approach:
Choose two people A and B.
Probability of A getting a seat at table is 1/2.
Probability of B getting a seat at the table once A has taken one is 3/7.
Probability of B getting a seat once A has taken one and sits opposite A is (1/3)(3/7) = 1/7.

So Answer = (1/2)(1/7) = 1/14

Is this approach correct?

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hey_thr67 Really wants to Beat The GMAT!
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Tue May 15, 2012 12:14 pm
A good question indeed. I think the correct answer should be 1/28.

Following is my solution.

Well for first person the probability to have seat is 1/4 ( as one seat is to be selected out of 4). Now, for the second person the probability of choosing the seat on square table is 1. But to the person selected out of rest 7 is 1/7.

Hence, total probability is (1/4) X (1) X ( 1/7) = 1/28

Cúchulainn Just gettin' started!
Joined
14 May 2012
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Tue May 15, 2012 12:57 pm
Thanks thr,

I got 1/2 for A, as there 8 people looking for a seat but only four seats, hence 4/8 probability of A getting a seat.

After A has taken a seat, there are 3 seats remaining with 7 people looking to sit down. So this would make a 3/7 probability of B getting a seat. Now only one of those 3 seats will be opposite A, so this 3/7 would need to be multiplied by 1/3 to get the probability of sitting opposite A.

So my final probability was got by multiplying all these probabilities: (1/2)(3/7)(1/3) = 1/14.

But again, I'm not sure if I'm missing something, so it would be good to get it confirmed whether this approach is right or not. Thanks again thr.

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