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A committee has ten members. There are two members that currently serve as the​ board's chairman and vice chairman. Each member is equally likely to serve in any of the positions. Two members are randomly selected and assigned to be the new chairman and vice chairman. What is the probability of randomly selecting the two members who currently hold the positions of chairman and vice chairman and reassigning them to their current​ positions?

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jmonterrozar

Answer:

1/90 = 1.11%

Step-by-step explanation:

We have that the number of ways of total selections and assignments possible is a permutation.

We know that permutations are defined like this:

nPr = n! / (n-r)!

In our case n = 10 and r = 2, replacing:

10P2 = 10! / (10 - 2)! = 10! / 8!

10P2 = 90

In addition to this, there will only be one way to randomly select the two members currently holding the positions of President and Vice President and reassign them to their current positions. Thus,

Probability would come being the following:

P = 1/90 = 1.11%

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