Drawing/Selecting/Picking Two/Three/More balls from a Box/Bag/Urn
Problem 3
Three balls are drawn at random from collection of 7 white, 12 green and 4 red balls. The probability that each is of different colour is
Solution
Total number of balls in the bag
| = | 7 White + 12 Green + 4 Red |
| = | 23 |
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 23 | |||
| ⇒ n | = | 23C3 | ||
|---|---|---|---|---|
| = |
| |||
| = | 1,771 |
Let A be the event of each of the balls drawn being of different colors/colours
For Event A [White] [Green] [Red] Total Available 7 12 4 23 To Choose 1 1 1 3 Choices 7C1 12C1 4C1 23C3
Number of Favorable Choices
= The number of ways in which the 3 balls each of a different
color/colour can be drawn from the total 23 = (Number of ways in which 1 white ball can be drawn from the total 7)
× (Number of ways in which 1 green ball can be drawn from the total 12)
× (Number of ways in which 1 red ball can be drawn from the total 4) = 7C1 × 12C1 × 4C1 = 7 × 12 × 4 = 336
Probability of each of the balls drawn being 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
= 336 1,771
= 48 253
• Odds
Number of Unfavorable Choices
= Total Number of possible choices − Number of Favorable choices ⇒ mAc = n − mA = 1,771 − 336 = 1,435
» in favor
Odds in Favor of each of the balls drawn being of different colors/colours
⇒ Odds in Favor of Event A
= Number of Favorable Choices : Number of Unfavorable Choices = mA : mAc = 336 : 1,435 = 48 : 205
» against
Odds against each of the balls drawn being of different colors/colours
⇒ Odds against Event A
= Number ofUnfavorable Choices : Number of Favorable Choices = mAc : mA = 1,435 : 336 = 253 : 48
| [White] | [Green] | [Red] | Total | |
|---|---|---|---|---|
| Available | 7 | 12 | 4 | 23 |
| To Choose | 1 | 1 | 1 | 3 |
| Choices | 7C1 | 12C1 | 4C1 | 23C3 |
Number of Favorable Choices
| = | The number of ways in which the 3 balls each of a different color/colour can be drawn from the total 23 |
| = | (Number of ways in which 1 white ball can be drawn from the total 7) × (Number of ways in which 1 green ball can be drawn from the total 12) × (Number of ways in which 1 red ball can be drawn from the total 4) |
| = | 7C1 × 12C1 × 4C1 |
| = | 7 × 12 × 4 |
| = | 336 |
Probability of each of the balls drawn being 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 |
|---|---|---|
| = | 1,771 − 336 | |
| = | 1,435 |
» in favor
Odds in Favor of each of the balls drawn being of different colors/colours
⇒ Odds in Favor of Event A
| = | Number of Favorable Choices : Number of Unfavorable Choices |
| = | mA : mAc |
| = | 336 : 1,435 |
| = | 48 : 205 |
» against
Odds against each of the balls drawn being of different colors/colours
⇒ Odds against Event A
| = | Number ofUnfavorable Choices : Number of Favorable Choices |
| = | mAc : mA |
| = | 1,435 : 336 |
| = | 253 : 48 |
Author : The Edifier