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Five individuals, including a and b, take seats around a circular table in a completely random fashion. suppose the seats are numbered 1, . . . , 5. 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?

Respuesta :

mathmate
Will use A and B in place of a and b for clarity.
Let x=number of individuals away from A, including A & B

Without loss of generality, assume A is seated in seat #1.

Then B is seated at 2,3,4,5 with equal probability.
Half of the time B is seated at 2 or 5, each of which is next to A, therefore x=2
The other half of the time B is seated at 3 or 4, each of which is separated from A by one seat, then x=3.

The expected number of individuals
E[X]=sum (x*P(x))
=2*(1/2)+3(1/2)
=2.5

So the expected number of individuals to handle the message is 2.5.

abiolataiwo2015

The number of  individuals you would expect to handle the message is 2.5.

Joint probability distribution

Let Z represent the number of individuals that handle the message

Table for the possible joint value of X and Y

Z                       Y

                         1          2          3            4         5  

X         1             -          2           3            3         2

          2           2         -           2            3         3

          3           3         2           -             2         3

           4           3         3           2            -          2

           5            2         3           3           2          -

Each cell contain=1/4×1/5=1/20

Hence:

Number of individual=10×2×1/20+10×3×1/20

Number of individual=20×0.05+30×0.05

Number of individual=2.5

Therefore the number of  individuals you would expect to handle the message is 2.5.

Learn more about Joint probability distribution here:https://brainly.com/question/17279418

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