Question

a committee has thirteen 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:

Calculate total possible assignments

Using the Combinatorics knowledge point

$$ P(13, 3) = \frac{13!}{(13-3)!} = 13 \times 12 \times 11 = 1716 $$

Determine the number of favorable outcomes

Using the Classical Probability knowledge point

$$ n(E) = 1 $$

Calculate the probability

Using the Classical Probability knowledge point

$$ P(E) = \frac{1}{1716} \approx 0.00058275 $$

Round to four decimal places

Using the Classical Probability knowledge point

$$ P(E) \approx 0.0006 $$

Answer:

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?

The probability is <blank>0.0006</blank>
(Round to four decimal places as needed.)