Drawing/Selecting/Picking Two/Three/More balls from a Box/Bag/Urn
Problem 6
From a bag containing 4 white and 5 black balls a man draws 3 at random. The odds against all the balls being black is
Solution
Total number of balls in the bag
| = | 4 White + 5 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 | ||
|---|---|---|---|---|
| = |
| |||
| = | 84 |
Let A be the event of all the balls drawn being black
For Event A [White] [Black] Total Available 4 5 9 To Choose 0 3 3 Choices 4C0 5C3 9C3
Number of Favorable Choices
= The number of ways in which the 3 black balls can be
drawn from the total 5 = 5C3 = 5 × 4 × 3 3 × 2 × 1
= 10
• Odds
Number of Unfavorable Choices
= Total Number of possible choices − Number of Favorable choices ⇒ mAc = n − mA = 84 − 10 = 74
» against
Odds against all the balls being black
⇒ Odds against Event A
= Number ofUnfavorable Choices : Number of Favorable Choices = mAc : mA = 74 : 10 = 37 : 5
| [White] | [Black] | Total | |
|---|---|---|---|
| Available | 4 | 5 | 9 |
| To Choose | 0 | 3 | 3 |
| Choices | 4C0 | 5C3 | 9C3 |
Number of Favorable Choices
| = | The number of ways in which the 3 black balls can be drawn from the total 5 | |||
| = | 5C3 | |||
| = |
| |||
| = | 10 |
• Odds
Number of Unfavorable Choices
| = | Total Number of possible choices − Number of Favorable choices | |
| ⇒ mAc | = | n − mA |
|---|---|---|
| = | 84 − 10 | |
| = | 74 |
» against
Odds against all the balls being black
⇒ Odds against Event A
| = | Number ofUnfavorable Choices : Number of Favorable Choices |
| = | mAc : mA |
| = | 74 : 10 |
| = | 37 : 5 |
Author : The Edifier