Drawing/Selecting/Picking Two/Three/More balls from a Box/Bag/Urn
Problem 4
A bag contains 3 red, 4 white and 5 blue balls, 2 balls are drawn at random. The probability that they are of different colours is
Solution
Total number of balls in the bag
| = | 3 Red + 4 White + 5 Blue |
| = | 12 |
Number of balls drawn = 2
Experiment : Drawing 2 balls from the bag
Total Number of Possible Choices
| = | Number of ways in which the two balls can be drawn from the total 12 | |||
| ⇒ n | = | 12C2 | ||
|---|---|---|---|---|
| = |
| |||
| = | 66 |
Let A be the event of drawing the balls such that they are of different colors/colours
For Event A Event A can be accomplished in three alternative ways. - A1 : Drawing 1 Red Ball and 1 White Ball
- A2 : Drawing 1 Red Ball and 1 Blue Ball
- A3 : Drawing 1 White Ball and 1 Blue Ball
[Red] [White] [Blue] Total Available 3 4 5 12 To Choose 1 1 0 2 A1 Choices 3C1 4C1 5C0 12C2 To Choose 1 0 1 2 A2 Choices 3C1 4C0 5C1 12C2 To Choose 0 1 1 2 A3 Choices 3C0 4C1 5C1 12C2
» "A1"
Number of Favorable Choices
= The number of ways in which a red ball and a white ball
can be drawn from the total 12 mA1 = (Number of ways in which 1 red ball can be drawn from the total 3)
× (Number of ways in which 1 white ball can be drawn from the total 4) = 3C1 × 4C1 = 3 × 4 = 12
» "A2"
Number of Favorable Choices
= The number of ways in which a red ball and a blue ball
can be drawn from the total 12 mA2 = (Number of ways in which 1 red ball can be drawn from the total 3)
× (Number of ways in which 1 blue ball can be drawn from the total 5) = 3C1 × 5C1 = 3 × 5 = 15
» "A3"
Number of Favorable Choices
= The number of ways in which a white ball and a blue ball
can be drawn from the total 12 mA3 = (Number of ways in which 1 white ball can be drawn from the total 4)
× (Number of ways in which 1 blue ball can be drawn from the total 5) = 4C1 × 5C1 = 4 × 5 = 20
Total number of Favorable/Favorable choices for Event A
⇒ mA = mA1 + mA2 + mA3 = 12 + 15 + 20 = 47
Probability of drawing the balls such that they are of different colors/colours
⇒ Probability of occurrence of Event A
= Number of Favorable Choices for the Event Total Number of Possible Choices for the Experiment
⇒ P(A) = mA n
= 47 66
• Odds
Number of Unfavorable Choices
= Total Number of possible choices − Number of Favorable choices ⇒ mAc = n − mA = 66 − 47 = 19
» in favor
Odds in Favor of drawing the balls such that they are of different colors/colours
⇒ Odds in Favor of Event A
= Number of Favorable Choices : Number of Unfavorable Choices = mA : mAc = 47 : 19
» against
Odds against drawing the balls such that they are of different colors/colours
⇒ Odds against Event A
= Number ofUnfavorable Choices : Number of Favorable Choices = mAc : mA = 19 : 47
- A1 : Drawing 1 Red Ball and 1 White Ball
- A2 : Drawing 1 Red Ball and 1 Blue Ball
- A3 : Drawing 1 White Ball and 1 Blue Ball
| [Red] | [White] | [Blue] | Total | ||
|---|---|---|---|---|---|
| Available | 3 | 4 | 5 | 12 | |
| To Choose | 1 | 1 | 0 | 2 | A1 |
| Choices | 3C1 | 4C1 | 5C0 | 12C2 | |
| To Choose | 1 | 0 | 1 | 2 | A2 |
| Choices | 3C1 | 4C0 | 5C1 | 12C2 | |
| To Choose | 0 | 1 | 1 | 2 | A3 |
| Choices | 3C0 | 4C1 | 5C1 | 12C2 |
» "A1"
Number of Favorable Choices
| = | The number of ways in which a red ball and a white ball can be drawn from the total 12 | |
| mA1 | = | (Number of ways in which 1 red ball can be drawn from the total 3) × (Number of ways in which 1 white ball can be drawn from the total 4) |
|---|---|---|
| = | 3C1 × 4C1 | |
| = | 3 × 4 | |
| = | 12 |
» "A2"
Number of Favorable Choices
| = | The number of ways in which a red ball and a blue ball can be drawn from the total 12 | |
| mA2 | = | (Number of ways in which 1 red ball can be drawn from the total 3) × (Number of ways in which 1 blue ball can be drawn from the total 5) |
|---|---|---|
| = | 3C1 × 5C1 | |
| = | 3 × 5 | |
| = | 15 |
» "A3"
Number of Favorable Choices
| = | The number of ways in which a white ball and a blue ball can be drawn from the total 12 | |
| mA3 | = | (Number of ways in which 1 white ball can be drawn from the total 4) × (Number of ways in which 1 blue ball can be drawn from the total 5) |
|---|---|---|
| = | 4C1 × 5C1 | |
| = | 4 × 5 | |
| = | 20 |
Total number of Favorable/Favorable choices for Event A
| ⇒ mA | = | mA1 + mA2 + mA3 |
|---|---|---|
| = | 12 + 15 + 20 | |
| = | 47 |
Probability of drawing the balls such that they are of different colors/colours
⇒ Probability of occurrence of Event A
| = |
| |||
| ⇒ P(A) | = |
| ||
|---|---|---|---|---|
| = |
|
• Odds
Number of Unfavorable Choices
| = | Total Number of possible choices − Number of Favorable choices | |
| ⇒ mAc | = | n − mA |
|---|---|---|
| = | 66 − 47 | |
| = | 19 |
» in favor
Odds in Favor of drawing the balls such that they are of different colors/colours
⇒ Odds in Favor of Event A
| = | Number of Favorable Choices : Number of Unfavorable Choices |
| = | mA : mAc |
| = | 47 : 19 |
» against
Odds against drawing the balls such that they are of different colors/colours
⇒ Odds against Event A
| = | Number ofUnfavorable Choices : Number of Favorable Choices |
| = | mAc : mA |
| = | 19 : 47 |
Author : The Edifier