Question

A committee has thirteen members. There are three members that currently serve as the​ board's 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?

Answer 1

Answer: Probability P = 1/1716

Step-by-step explanation:

Definition

For permutation ( order is important)

nPr = n!/(n-r)!

Given;

Total number of members in the committee = 13

Total number of members to be selected = 3

Since order is important in this case.

The total number (Tt) of possible ways of selecting three executives from the committee members is given as

Tt = 13P3 (since order is important)

Tt = 13!/(13-3)!

Tt = 13!/10!

Tt = 1716

The probability P of randomly selecting the three members who currently hold the positions of chairman, ranking member, and treasurer and reassigning them to their current​ positions is equal to

P = number of favourable outcomes/ number of possible outcomes

Number of favourable outcomes = 1 (i.e 1×1×1 = 1) one option for each post.

Number of possible outcomes = Tt = 1716

P = 1/1716