Question

Ex. 24

Six individuals, including A and B, take seats around a circular table in a completely random fashion. Suppose the seats are numbered 1, …, 6. Let X= A's seat number and Y= B's seat number. If A sends a written message around the table to B in the direction in which they are closest, how many individuals (including A and B) would you expect to handle the message?

Answer 1

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From the above figure it is evident there are minimum 0 and maximum 2 people can sit between A and B.

So possible are 0,1,2 people sitting between A and B

As this is a case of circular permutations, n people randomly be arranged around a circular table in (n-1)! ways

Case 1: Probability that no individual sits between A and B

There are a total of 6 members. Since 2 particular people always together, consider them as one person say P. Now, P with other 4, so 5 can be arranged around the table in 4! ways. The two (A and B) between themselves can be arranged in 2! ways.

Hence the required probability is 4! 2!/5! = 2/5

Case 2: Probability that 1 individual sits between A and B

First, select one person from remaining 4 (i.e. excluding A and B) = 4c1

Now arrange this person along with A and B as a group say P. Now, P with other 3, so 4 can be arranged around the table in 3!. Also, the two (A and B) between themselves can be arranged in 2! ways.

Hence the required probability is 4c1*2!*3!/5! = 2/5

Case 2: Probability that 2 individual sits between A and B

First, select two persons from the remaining 4 (i.e. excluding A and B) = 4c2 ways.

Now arrange this person along with A and B as a group say P. Now, P with other 2, so 3 can be arranged around the table in 2!. Also, the two (A and B) between themselves can be arranged in 2! ways. Also the initial randomly selected 2 persons between themselves can be arranged in 2! ways

Hence the required probability is 4c2*2!*2!/5! = 1 /5

Now Expected number of people sitting between A and B is given by

E(X) = n1*p1 + n2*p2 + n3*p3

       = 0*(2/5) + 1*(2/5) + 2*(1/5)

E(X) = 4/5 = 0.8

So expected number of people sitting between A and B is 0.8