Question

A committee has members. There are members that currently serve as the​ board's . Each member is equally likely to serve in any of the positions. members are randomly selected and assigned to be the new . What is the probability of randomly selecting the members who currently hold the positions of and reassigning them to their current​ positions? The probability is nothing.

Answer 1

Complete Question

A committee has six members. There are three members that currently serve as the​ board's chairman comma vice chairman comma 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 comma vice chairman comma and treasurer . What is the probability of randomly selecting the three members who currently hold the positions of chairman comma vice chairman comma and treasurer and reassigning them to their current​ positions?

The probability is ?

Answer:

The probability is $P(3 ) = \frac{ 1 }{20 }$

Step-by-step explanation:

From the question we are told that

       The total number of members is  n = 6  

       The  number of member to be selected is  r =  3  

Generally the number of ways of selecting 3 members from 6  is mathematically evaluated as

          $\left n } \atop {}} \right. C _r = \frac{n! }{(n-r ) ! r !}$

=>       $\left 6 } \atop {}} \right. C _ 3 = \frac{6 ! }{(6-3) ! 3 !}$

=>        $\left n } \atop {}} \right. C _r = \frac{6* 5* 4 * 3! }{3 ! 3*2 * 1}$

=>       $\left n } \atop {}} \right. C _r = \frac{6* 5* 4 }{3 *2 * 1}$

=>        $\left n } \atop {}} \right. C _r =20$

Now the number of ways of selecting the 3 members who currently hold the position is  n  = 1

  So the probability is mathematically represented as

                 $p(k ) = \frac{ n }{\left n } \atop {}} \right. C _r }$

substituting values

                  $P(3 ) = \frac{ 1 }{20 }$