Drawing/Selecting/Picking/Choosing Two/Three/More fruits/lamps/bulbs/objects/things from a Box/Bag/Urn
Problem : 1
Three mangoes and three apples are in a box. If two fruits are chosen at random. The probability that one is mango and the other is an apple is
Solution
Total number of fruits in the box
| = | three mangoes + three apples |
| = | 6 fruits |
Number of fruits selected = 2
Experiment : Picking 2 fruits from the box
Total Number of Possible Choices
| = | Number of ways in which the two fruits can be drawn from the total 6 fruits | |||
| ⇒ n | = | 6C2 | ||
|---|---|---|---|---|
| = |
| |||
| = | 15 |
Let A be the event of the two fruits being one a mango and the other an apple
For Event A [Mangoes] [Apples] Total Available 3 3 6 To Choose 1 1 2 Choices 3C1 3C1 6C2
Number of Favorable Choices
= The number of ways in which a mango and an apple
can be selected from the total 6 fruits ⇒ mA = (Number of ways in which one mango can be selected from the 3)
× (Number of ways in which one apple can be selected from the 3) 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.
= 3C1 × 3C1 = 3 × 3 = 9
Probability of the two fruits being one a mango and the other an apple
⇒ 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
= 9 15
= 3 5
• Odds
Number of Unfavorable Choices
= Total Number of possible choices − Number of Favorable choices ⇒ mAc = n − mA = 15 − 9 = 6
» in favor
Odds in Favor of the two fruits being one a mango and the other an apple
⇒ Odds in Favor of Event A
= Number of Favorable Choices : Number of Unfavorable Choices = mA : mAc = 9 : 6 = 3 : 2
» against
Odds against the two fruits being one a mango and the other an apple
⇒ Odds against Event A
= Number ofUnfavorable Choices : Number of Favorable Choices = mAc : mA = 6 : 9 = 2 : 3
| [Mangoes] | [Apples] | Total | |
|---|---|---|---|
| Available | 3 | 3 | 6 |
| To Choose | 1 | 1 | 2 |
| Choices | 3C1 | 3C1 | 6C2 |
Number of Favorable Choices
| = | The number of ways in which a mango and an apple can be selected from the total 6 fruits | ||
| ⇒ mA | = | (Number of ways in which one mango can be selected from the 3) × (Number of ways in which one apple can be selected from the 3)
| |
|---|---|---|---|
| = | 3C1 × 3C1 | ||
| = | 3 × 3 | ||
| = | 9 |
Probability of the two fruits being one a mango and the other an apple
⇒ 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 |
|---|---|---|
| = | 15 − 9 | |
| = | 6 |
» in favor
Odds in Favor of the two fruits being one a mango and the other an apple
⇒ Odds in Favor of Event A
| = | Number of Favorable Choices : Number of Unfavorable Choices |
| = | mA : mAc |
| = | 9 : 6 |
| = | 3 : 2 |
» against
Odds against the two fruits being one a mango and the other an apple
⇒ Odds against Event A
| = | Number ofUnfavorable Choices : Number of Favorable Choices |
| = | mAc : mA |
| = | 6 : 9 |
| = | 2 : 3 |
Author : The Edifier