Question

a committee has ten members. there are three members that currently serve as the boards chairman, ranking member, and treasurer. each member is equally likely to serve in any of the positions. three members are randomly selected and assigned to be the new chairman, ranking member, and treasurer. what is the probability of randomly selecting the three members who currently hold the positions of chairman, ranking member, and treasurer and reassigning them to their current positions?
the probability is
(round to four decimal places as needed.)

Explanation:

Step1: Calculate the total number of permutations

The number of ways to select and assign 3 members out of 10 for 3 distinct positions (chairman, ranking member, treasurer) is given by the permutation formula \(P(n,r)=\frac{n!}{(n - r)!}\), where \(n = 10\) and \(r=3\).

$$P(10,3)=\frac{10!}{(10 - 3)!}=\frac{10!}{7!}=10\times9\times8 = 720$$

Step2: Calculate the number of favorable permutations

There is only 1 way to re - assign the 3 current members (chairman, ranking member, treasurer) to their current positions.

Step3: Calculate the probability

The probability \(P\) is the number of favorable outcomes divided by the number of total outcomes. So \(P=\frac{1}{720}\approx0.0014\)

Answer:

\(0.0014\)