Question

This question concerns forming committees of 4 persons from a group of 10 people.

(a) How many ways are there to choose a committee of 4 persons from a group of 10 persons, if order is not important?

(b) If each of these outcomes from part (a) is equally likely then what is the probability that a particular person is on the committee?

(c) What is the probability that a particular person is not on the committee?

(d) How many ways are there to choose a committee of 4 persons from a group of 10 persons, if one is to be the chairperson?

Answer 1

One of the basics of counting principle is that if we have n distinct objects and we need to choose k objects out of these n distict objects ,not considering their order of selection, then the number of ways in which this can be done is

, read as n combination k or simply "n choose k".

(a) Required number of ways to choose a committee of 4 persons from a group of 10 persons, order being not important is :

= 210

(b) We first consider the number of ways in which a particular person is on the committee. To do this, we keep aside that particular person (as he is already selected) and just choose 3 persons out of the remaining 9 persons.

This can be done in   ways i,.i.e., 84 ways

As, all possible outcomes (the value obtained in part (a)) are equally likely,

Required probability = No of favourable outcomes / Total number of outcomes

= 84/210

(c) We first find the number of ways a particular person is not on the committee. To do this, we first throw that person away, and choose all the 4 persons out of the remaining 9 persons.

This can be done in ways ,i.e. 126 ways.

Using similar argument as in part (b),

Required probability

= 126/210

(d) We first choose 4 persons out of 10 persons in ways and corresponding to each such selection, we choose the chairperson among the 4 selected persons in ways.

Required number of ways=

=210*4

=840