Problem 1
The probability of getting 2 white and 3 black balls from a bag containing 8 white and 12 black balls is
Solution
Total number of balls in the bag
| = | 8 White + 12 Black |
| = | 20 |
Number of balls drawn = 5
Experiment : Drawing 5 balls from the bag
Total Number of Possible Choices
| = | Number of ways in which five balls can be drawn from the total 20 |
| ⇒ n | = | 20C5 |
|---|
| = | | 20 × 19 × 18 × 17 × 16 | | 5 × 4 × 3 × 2 × 1 |
|
| = | 19 × 6 × 17 × 8 |
| = | 15,504 |
Let A be the event of drawing 2 white and 3 black balls
For Event A | [White] | [Black] | Total |
|---|
| Available | 8 | 12 | 20 |
|---|
| To Choose | 2 | 3 | 5 |
|---|
| Choices | 8C2 | 12C3 | 20C5 |
|---|
Number of Favorable Choices
| = | The number of ways in which the 2 white and 3 black balls can be
drawn from the total 20 |
| ⇒ mA | = | (Number of ways in which the 2 white balls can be drawn from the total 8)
× (Number of ways in which the 3 black balls can be drawn from the total 12) |
|---|
| = | 8C2 × 12C3 |
| = | |
| = | 28 × 220 |
| = | 6,160 |
Probability of drawing 2 white and 3 black balls
⇒ Probability of occurrence of Event A
| = | | Number of Favorable Choices for the Event | | Total Number of Possible Choices for the Experiment |
|
| ⇒ P(A) | = | |
|---|
| = | |
| = | |
• Odds
Number of Unfavorable Choices
| = | Total Number of possible choices − Number of Favorable choices |
| ⇒ mAc | = | n − mA |
|---|
| = | 15,504 − 6,160 |
| = | 9,344 |
» in favor
Odds in Favor of drawing 2 white and 3 black balls
⇒ Odds in Favor of Event A
| = | Number of Favorable Choices : Number of Unfavorable Choices |
| = | mA : mAc |
| = | 6,160 : 9,344 |
| = | 385 : 584 |
» against
Odds against drawing 2 white and 3 black balls
⇒ Odds against Event A
| = | Number ofUnfavorable Choices : Number of Favorable Choices |
| = | mAc : mA |
| = | 9,344 : 6,160 |
| = | 584 : 385 |
