Drawing/Selecting/Picking Two/Three/More balls from a Box/Bag/Urn

Define the Event and identify the number of favourable choices in the following cases which relate to the experiment of drawing balls

a)	The probability of drawing 3 white and 4 green balls from a bag containing 5 white and 6 green balls if the seven balls are drawn at random simultaneously
b)	Three balls are drawn from a bag containing 10 white and 15 red and 5 green balls. Find the probability that a) They are white b) They are red c) They are green
c)	A box contains 2 red, 3 blue and 4 black balls. Three black are drawn from the box at random. What is the probability that (i) the three balls are of different colours? (ii) two balls are of the same colour and the third of different colour? (iii) all the balls are of the same colour?

Total number of balls in the bag

=	5 White + 6 Green
=	11

Number of balls drawn = 7

Experiment : Drawing 7 balls from the bag

Total Number of Possible Choices

=

Number of ways in which the seven balls can be drawn from the total 11

⇒ n

=

¹¹C₇

=

¹¹C_{(11 − 7)} ⁿC_r = ⁿC_{(n − r)}

=

¹¹C₄

=

11 × 10 × 9 × 8

4 × 3 × 2 × 1

=

330

Let A be the event of drawing 3 white and 4 green balls

For Event A

	[White]	[Green]	Total
Available	5	6	11
To Choose	3	4	7
Choices	⁵C₃	⁶C₄	¹¹C₇

Number of Favorable Choices

=

The number of ways in which the 3 whtie and 4 green balls can be
drawn from the total 11

m_A

=

(Number of ways in which the 3 white balls can be drawn from the total 5)
× (Number of ways in which the 4 green balls can be drawn from the total 6)

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.

=

⁵C₃ × ⁶C₄

=

⁵C_{(5 − 3)} × ⁶C_{(6 − 4)}

=

⁵C₂ × ⁶C₂

=

5 × 4

2 × 1

×

6 × 5

2 × 1

=

10 × 15

=

150

Total number of balls in the bag

=	10 White + 15 Red + 5 Green
=	30

Number of balls drawn = 3

Experiment : Drawing 3 balls from the bag

Total Number of Possible Choices

=

Number of ways in which the seven balls can be drawn from the total 11

⇒ n

=

³⁰C₃

=

30 × 29 × 28

3 × 2 × 1

=

4,060

Let

A be the event of the three balls being white
B be the event of the three balls being red
C be the event of the three balls being green

For Event A

	[White]	[Red]	[Green]	Total
Available	10	15	5	30
To Choose	3	0	0	3
Choices	¹⁰C₃	¹⁵C₀	⁵C₀	³⁰C₃

Number of Favorable Choices

=

The number of ways in which the 3 white balls can be drawn
from the total 30

m_A

=

Number of ways in which the 3 white balls can be drawn from the total 10

=

¹⁰C₃

=

10 × 9 × 8

3 × 2 × 1

=

120

For Event B

	[White]	[Red]	[Green]	Total
Available	10	15	5	30
To Choose	0	3	0	3
Choices	¹⁰C₀	¹⁵C₃	⁵C₀	³⁰C₃

Number of Favorable Choices

=

The number of ways in which the 3 red balls can be drawn
from the total 30

m_A

=

Number of ways in which the 3 red balls can be drawn from the total 15

=

¹⁵C₃

=

15 × 14 × 13

3 × 2 × 1

=

455

For Event C

	[White]	[Red]	[Green]	Total
Available	10	15	5	30
To Choose	0	0	3	3
Choices	¹⁰C₀	¹⁵C₀	⁵C₃	³⁰C₃

Number of Favorable Choices

=

The number of ways in which the 3 green balls can be drawn
from the total 30

m_A

=

Number of ways in which the 3 green balls can be drawn from the total 5

=

⁵C₃

=

5 × 4 × 3

3 × 2 × 1

=

10

Total number of balls in the bag

=	2 Red + 3 Blue + 4 Black
=	9

Number of balls drawn = 3

Experiment : Drawing 3 balls from the bag

Total Number of Possible Choices

=

Number of ways in which the three balls can be drawn from the total 9

⇒ n

=

⁹C₃

=

9 × 8 × 7

3 × 2 × 1

=

84

Let

"P" be the event of the three balls being of different colors/colours
"Q" be the event of two balls being of the same color/colour and the third of a different color/colour
"R" be the event of the three balls being of the same color/colour

For Event "P"

	[Red]	[Blue]	[Black]	Total
Available	2	3	4	9
To Choose	1	1	1	3
Choices	²C₁	³C₁	⁴C₁	⁹C₃

Number of Favorable Choices

=

The number of ways in which the 3 balls of different colors/colours
can be drawn from the total 9

m_P

=

(Number of ways in which 1 red ball can be drawn from the total 2)
× (Number of ways in which 1 blue ball can be drawn from the total 3)
× (Number of ways in which 1 black ball can be drawn from the total 4)

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.

=

²C₁ × ³C₁ × ⁴C₁

=

2 × 3 × 4

=

24

For Event "Q"

Fundamental Counting Theorem (of Addition): Where an event can be accomplished in a number of alternative ways, the total number of ways in which the event can be accomplished is equal to the sum of the number of ways in which the alternative events can be accomplished.

Event "Q" can be accomplished in 6 alternative ways

Q₁ : Drawing 2 Red balls and 1 Blue ball
Q₂ : Drawing 2 Red balls and 1 Black ball
Q₃ : Drawing 2 Blue balls and 1 Red ball
Q₄ : Drawing 2 Blue balls and 1 Black ball
Q₅ : Drawing 2 Black balls and 1 Red ball
Q₆ : Drawing 2 Blue balls and 1 Black ball

	[Red]	[Blue]	[Black]	Total
Available	2	3	4	9
To Choose	2	1	0	3	Q₁
Choices	²C₂	³C₁	⁴C₀	⁹C₃	Q₁
To Choose	2	0	1	3	Q₂
Choices	²C₂	³C₀	⁴C₁	⁹C₃	Q₂
To Choose	1	2	0	3	Q₃
Choices	²C₁	³C₂	⁴C₀	⁹C₃	Q₃
To Choose	2	0	1	3	Q₄
Choices	²C₂	³C₀	⁴C₁	⁹C₃	Q₄
To Choose	1	0	2	3	Q₅
Choices	²C₁	³C₀	⁴C₂	⁹C₃	Q₅
To Choose	0	1	2	3	Q₆
Choices	²C₀	³C₁	⁴C₂	⁹C₃	Q₆

Total number of Favorable/Favorable choices for Event "Q"

⇒ m_Q

=

m_Q1 + m_Q2 + m_Q3 + m_Q4 + m_Q5 + m_Q6

=

(²C₂ × ³C₁ × ⁴C₀) + (²C₂ × ³C₀ × ⁴C₁) + (²C₁ × ³C₂ × ⁴C₀)
+ (²C₀ × ³C₂ × ⁴C₁) + (²C₁ × ³C₀ × ⁴C₂) + (²C₀ × ³C₁ × ⁴C₂)

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.

=

(1 × 3 × 1) + (1 × 1 × 4) + (2 × 3 × 1) + (1 × 3 × 4) + (2 × 1 × 6) + (1 × 3 × 6)

=

3 + 4 + 6 + 12 + 12+ 18

=

55

For Event "R"

Fundamental Counting Theorem (of Addition): Where an event can be accomplished in a number of alternative ways, the total number of ways in which the event can be accomplished is equal to the sum of the number of ways in which the alternative events can be accomplished.

Event "R" can be accomplished in 2 alternative ways

R₁ : Drawing 3 Blue ball
R₂ : Drawing 3 Black ball

	[Red]	[Blue]	[Black]	Total
Available	2	3	4	9
To Choose	0	3	0	3	R₁
Choices	²C₀	³C₃	⁴C₀	⁹C₃	R₁
To Choose	0	0	3	3	R₂
Choices	²C₀	³C₀	⁴C₃	⁹C₃	R₂

Total number of Favorable/Favorable choices for Event "R"

⇒ m_R

=

m_R1 + m_R2

=

(²C₀ × ³C₃ × ⁴C₀) + (²C₀ × ³C₀ × ⁴C₃)

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.

=

(1 × 1 × 1) + (1 × 1 × 4)

=

1 + 4

=

5

Drawing/Selecting/Picking Two/Three/More balls from a Box/Bag/Urn

Problem 7

Solution a

For Event A

Solution (b)

For Event A

For Event B

For Event C

Solution (c)

For Event "P"

For Event "Q"

For Event "R"