Question

The patio of a restaurant has three tables at which two, four, and five patrons can be seated, respectively. A party of eleven is to be seated on the patio.

(a) In how many ways can the party of eleven be seated at the three tables (it is not important in which seats at each table the patrons sit).

(b) Three friends among the party wish to sit together. In how many ways can the party of eleven be seated such that the three friends sit at the same table?

(c) Suppose a seating arrangement is randomly selected among all possible seating arrangements such that each arrangement is equally likely. Find the probability that the three friends are seated at the same table.

(d) Suppose one of the three friends arrives early and seats herself at the table at which five patrons can be seated. Supposing that the remaining members of the party are assigned seats at random, find the probability that the three friends end up sitting at the same table.

(e) Find the probability that the three friends end up sitting at the same table given that the friend arriving early sits at the table at which four patrons can be seated.

Answer 1

P(Event) = Number of favorable outcomes/Total Number of outcomes

Please note

Assuming that these are round tables, therefore the number of ways n people can be seated = (n - 1)!

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(a) 2 members can be chosen for the 2 chair table in 11C2 = 11 * 10 / 2 = 55 ways.

These 2 can be arranged in (2 - 1)! = 1 way

Therefore Number of ways of arrangements for the 2 member table = 55 * 2 = 110

Remaining are 9 members, of which 4 can be chosen 9C4 = 9 * 8 * 7 * 6 / 4 * 3 * 2 * 1 = 126 ways and can be seated in (4 - 1)! = 3! = 6 ways

Therefore Number of ways of arrangements for the 4 member table = 126 * 6 = 756

Remaining are 5 members, of which all 5 can be chosen 5C5 = 1 way and can be seated in (5 - 1)! = 4! = 24 ways

Therefore the total number of ways 110 * 756 * 24 = 1995840 ways

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(b) The three friends can sit together either at the 4 member table or 5 member table

Suppose they are seated at the 4 member table

Since these 3 are out, there are 8 members remaining. Now 1 member can be chosen for the 4 member table in 8C1 = 8 ways. These 4 members can be seated in (4 - 1)! = 3! = 6 ways

Therefore Total number of ways of seating the 4 members = 8 * 6 = 48

Remaining are 7 members, of which 2 can be chosen in 7C2 = 7 * 6 / 2 * 1 = 21 ways and can be seated in (2 - 1)! = 1! = 1 way

Therefore Number of ways of arrangements for the 4 member table = 21 * 1 = 21

Remaining are 5 members, of which all 5 can be chosen 5C5 = 1 way and can be seated in (5 - 1)! = 4! = 24 ways

Therefore the total number of ways 48 * 21 * 24 = 24192 ways ---- (1)

Suppose they are seated at the 5 member table

Since these 3 are out, there are 8 members remaining. Now 2 members can be chosen for the 5 member table in 8C2 = 28 ways. These 5 members can be seated in (5 - 1)! = 4! = 24 ways

Therefore Total number of ways of seating the 4 members = 28 * 24 = 672 ways

Remaining are 6 members, of which 2 can be chosen in 6C2 = 6 * 5 / 2 * 1 = 15 ways and can be seated in (2 - 1)! = 1! = 1 way

Therefore Number of ways of arrangements for the 4 member table = 15 * 1 = 15

Remaining are 4 members, of which all 4 can be chosen 4C4 = 1 way and can be seated in (4 - 1)! = 3! = 6 ways

Therefore the total number of ways 672 * 15 * 6 = 60480 ways ---- (2)

Therefore the total number of ways the three friends sit together = 24192 + 60480 = 84672 ways

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(c) Probability = Favorable Outcomes / Total Outcomes

Favorable outcomes = Outcomes found in (b) = 84672

Total Outcomes = Outcomes found in (a) = 1995840

Therefore the required probability = 84672 / 1995840 = 0.0424

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(d) Probability = Favorable Outcomes / Total Outcomes

Favorable outcomes = When all three friends sit at the 5 member table = 60480

Total Outcomes = Outcomes found in (a) = 1995840

Therefore the required probability = 60480 / 1995840 = 0.0303

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(e) By Bayes Theorem, P(A given B) = P(A and B) / P(B)

P(All sit together given that the friend arriving early sits at the 4 member table)

= P(All sit together and the friend arriving early sits at the 4 member table) / P(The friend arriving early, sits at the 4 member table)

P(All sit together and the friend arriving early sits at the 4 member table) = From (b) the number of ways = 24192 and therefore the probability = 24192/1995840

P(The friend arriving early sits at the 4 patron table)

Favorable Outcomes :

Since this 1 person out, there are 10 members remaining. Now 3 members can be chosen for the 4 member table in 10C3 = 120 ways. These 4 members can be seated in (4 - 1)! = 3! = 6 ways

Therefore Total number of ways of seating the 4 members = 120 * 6 = 720

Remaining are 7 members, of which 2 can be chosen in 7C2 = 7 * 6 / 2 * 1 = 21 ways and can be seated in (2 - 1)! = 1! = 1 way

Therefore Number of ways of arrangements for the 4 member table = 21 * 1 = 21

Remaining are 5 members, of which all 5 can be chosen 5C5 = 1 way and can be seated in (5 - 1)! = 4! = 24 ways

Therefore the total number of ways 120 * 21 * 24 = 60480 ways

Total Outcomes = Outcomes found in (a) = 1995840

P(The person who came early sits on the table for 4) = 60480 / 1995840

Therefore the required probability = (24192/1995840) / (60480/1995840) = 24192 / 60480 = 0.4

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