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Arsal

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Probability

If k people are seated in a random manner in a row containing n seats (n>k), what is the probability that the peoplewill occupy k adjacent seats in the row?

luca-deltodesco

Member

Registered: 2006-05-05

Posts: 1,470

Re: Probability

first, lets see the TOTAL number of possibilities for arrangements, this will be nPk. (permutations)
now, the number of possibilities for all k people to be in the first k seats of n chairs will be k!.
now, the number of sets of k adjacent seats in n chairs will be n-k+1.

so the probability that k people will be sat adjacent in a row of n charis, is k!(n-k+1)/nPk

nPk = n!/(n-k)! i.e.

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John E. Franklin

Member

Registered: 2005-08-29

Posts: 3,588

Re: Probability

Test with numbers:
5 people and 7 seats.
5 people sit on the left with two empty seats on the right.
5 people sit in the middle with an empty seat on both ends.
5 people sit on the right with two empty seats on the left.
Number people 1 to 5: 1 2 3 4 5 and number the empty seats 6 and 7.
Now the 6 and 7 empty seats are filled with emptiness, which is equal, so 6 and 7 are equal.
So 7 6 and 6 7 order is just one combination.
...

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igloo myrtilles fourmis

John E. Franklin

Member

Registered: 2005-08-29

Posts: 3,588

Re: Probability

But however, if two more people are added and named empty6 and empty7, now the two empties are different!
This must changed the original problem which should be reread now.

Probability

If 5 people are seated in a random manner in a row containing 7 seats (7>5),
what is the probability that the people will occupy 5 adjacent seats in the row?

So when the seating occurs, there are only 5 people, and if we now add two
people with the names empty 6 and empty 7, this causes the sitting process to
take longer and increases the number of combinations, since now 6 and 7 are
distinguishable.
...

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igloo myrtilles fourmis

John E. Franklin

Member

Registered: 2005-08-29

Posts: 3,588

Re: Probability

If the imaginary person named empty6 sits down after 1 and 2 sit, but before 3, 4, and 5, and empty7 sit, then we have a
new scenario, that didn't exist before.  When 3 looks at his options to sit down, he must avoid sitting on empty6 now!
So the randomness has changed.
...

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igloo myrtilles fourmis

John E. Franklin

Member

Registered: 2005-08-29

Posts: 3,588

Re: Probability

Since we don't want to upset the randomness, we make a rule that the imaginary empty6 and 7 people have to go last!!
So now it's like they never existed in the first place, since the if they go 6 7 or 7 6 in the end, we don't care anyway.

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igloo myrtilles fourmis

luca-deltodesco

Member

Registered: 2006-05-05

Posts: 1,470

Re: Probability

wait, what are you on about john?

ill take youre first post, 5 people, 7 seats

ignore the individual placements, first think of the 5 adjacent seats that can be taken, you can have

ooooo--
-ooooo-
--ooooo  3 possible sets of 5 adjacent seats

for each of those sets, the 5 people can sit down in any order, and the total number of placements for the 5 people, is 5!
so the total possible arrangements possible to satisfy the condition is 5! x 3. or k!(n-k+1)

then you have to think of the total number of ways that people can sit, which is the number of permutations. n!/(n-k)!

for example, for a set of 4 seats, 3 people can sit like

abc-, acb-, bac-, bca-, cab-, cba-
a-bc,a-cb,b-ac,b-ca,c-ab,c-ba
ab-c,ac-b,ba-c,bc-a,ca-b,cb-a
-abc, -acb, -bac, -bca, -cab, -cba

which is equal to 4P3 (24)

given the total number of possible arrangements, and the total number of possible arrangements satisfying the condition, you can simply find the probability that an arrangement will meet the condition, by dividing the satisfying number by total number

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