|
In how many ways can a committee 4 members be chosen from a group of 6 men and 5 women such that the committee consists of at least 2 women?
|
| Total number of members |
= |
6 men + 5 women |
|
= |
11 |
Number of Members selected = 4.
Experiment : Choosing 4 members
Let "E" be the event of choosing the committee with at least 2 women in it.
• For Event E
Event E can be accomplished in three alternative ways
|
Choosing the Committee with 4 members
|
| Ea » Choosing the Committee with 2 Women and 2 Men |
| Eb » Choosing the Committee with 3 Women and 1 Man |
| Ec » Choosing the Committee with 4 Women and 0 Men |
- Total Event (E) = Choosing the committee of 4 members with at least 2 women in it.
- 1st alternative-event (Ea) = Choosing the committee with 2 women and 2 men
- 2nd alternative-event (Eb) = Choosing the committee with 3 women and 1 man
- 3rd alternative-event (Ec) = Choosing the committee with 4 women and 0 men
[Ea, Eb, Ec are Mutually Exclusive Events. Occurrence of one of these events prevents the occurrence of the others. If the committee is chosen in one of these ways we can say that it was not chosen in the other ways]
» Ea
|
The number of ways in which the committee can be chosen with 2 women and 2 men
|
= |
150 |
| n(Ea) |
= |
150 |
Working Notes :
Considering the event of choosing the 2 women and 2 men to be the total event.
The total event can be sub-divided into two independent sub-events
- Total Event (Ea) = Choosing the 2 women and 2 men
- 1st sub-event (Ea1) = Choosing the 2 women from the available 6
- 2nd sub-event (Ea2) = Choosing the 2 men from the available 5
| Working Table |
|
Women |
× |
Men |
| Available |
6 |
5 |
| To Choose |
2 |
2 |
| Choices |
6C2 |
5C2 |
|
|
Fundamental Counting Theorem (of Multiplication): Where an event can be sub divided into two or more independent sub-events, the total number of ways in which the total event can be accomplished is equal to the product of the number of ways in which the sub-events can be accomplished.
|
|
| By the fundamental counting theorem of multiplication, nEa |
= |
nEa1 × nEa2 |
|
= |
6C2 × 5C2 +
|
|
= |
|
|
= |
15 × 10 |
|
= |
150 |
» Eb
|
The number of ways in which the committee can be chosen with 3 women and 1 man
|
= |
100 |
| n(Eb) |
= |
150 |
Working Notes :
Considering the event of choosing the 3 women and 1 man to be the total event.
The total event can be sub-divided into two independent sub-events
- Total Event (Eb) = Choosing the 3 women and 1 man
- 1st sub-event (Eb1) = Choosing the 3 women from the available 6
- 2nd sub-event (Eb2) = Choosing the 1 man from the available 5
| Working Table |
|
Women |
× |
Men |
| Available |
6 |
5 |
| To Choose |
3 |
1 |
| Choices |
6C3 |
5C1 |
|
|
Fundamental Counting Theorem (of Multiplication): Where an event can be sub divided into two or more independent sub-events, the total number of ways in which the total event can be accomplished is equal to the product of the number of ways in which the sub-events can be accomplished.
|
|
| By the fundamental counting theorem of multiplication, nEb
|
= |
nEb1 × nEb2 |
|
= |
6C3 × 5C1
|
|
= |
|
|
= |
20 × 5 |
|
= |
100 |
» Ec
|
The number of ways in which the committee can be chosen with 4 women and 0 men
|
= |
15 |
| n(Ec) |
= |
15 |
Working Notes :
Considering the event of choosing the 4 women and 0 men to be the total event.
The total event can be sub-divided into two independent sub-events
- Total Event (Ec) = Choosing the 4 women and 0 men
- 1st sub-event (Ec1) = Choosing the 4 women from the available 6
- 2nd sub-event (Ec2) = Choosing the 0 men from the available 5
| Working Table |
|
Women |
× |
Men |
| Available |
6 |
5 |
| To Choose |
4 |
0 |
| Choices |
6C4 |
5C0 |
|
|
Fundamental Counting Theorem (of Multiplication): Where an event can be sub divided into two or more independent sub-events, the total number of ways in which the total event can be accomplished is equal to the product of the number of ways in which the sub-events can be accomplished.
|
|
| By the fundamental counting theorem of multiplication, nEc |
= |
nEc1 × nEc2 |
|
= |
6C4 × 5C0
|
|
= |
| 6 × 5 × 4 × 3 | | 4 × 3 × 2 × 1 |
|
× |
1 |
|
|
= |
15 × 1 |
|
= |
15 |
Therefore,
The number of ways in which the committee can be chosen
= [No. of ways in which the committee with 2 women and 2 men (1st alternative event) can be chosen]
+ [No. of ways in which the committee with 3 women and 1 man (2nd alternative-event) can be chosen]
+ [No. of ways in which the committee with 4 women and 0 men (3rd alternative-event) can be chosen]
|
⇒ nE
|
= |
nEa + nEb + nEc
|
|
= |
150 + 100 + 15 |
|
= |
265 |