Drawing/Selecting/Picking/Choosing Two/Three/More fruits/lamps/bulbs/objects/things from a Box/Bag/Urn
Problem 3
In a box containing 15 bulbs, 5 are defective. If 5 bulbs are selected from the box, at random then find the probability of the event that i) none of them is defective ii) only one of them is defective iii) At least one of them is defective.
Solution
Total number of bulbs in the box
| = | 5 Defective + 10 Non-Defective |
| = | 15 |
Number of bulbs selected = 5
Experiment : Selecting 5 bulbs from the box
Total Number of Possible Choices
| = | Number of ways in which the five bulbs can be selected from the total 15 bulbs | |||
| ⇒ n | = | 15C5 | ||
|---|---|---|---|---|
| = |
| |||
| = | 7 × 13 × 3 × 11 | |||
| = | 3,003 |
Let
- A be the event of none of the bulbs being defective
- B be the event of only one of the bulbs being defective
- C be the event of at least one of the bulbs being defective
For Event A None of the bulbs being a defective ⇒ All the bulbs are non-defective. [Defectives] [Non-Defectives] Total Available 5 10 15 To Choose 0 5 5 Choices 5C0 10C5 15C5
Number of Favorable Choices
= The number of ways in which 5 non-defective bulbs can be
selected from the total 15 bulbs ⇒ mA = (Number of ways in which 5 non-defective bulbs can be selected from the 10) = 10C5 = 10 × 9 × 8 × 7 × 6 5 × 4 × 3 × 2 × 1
= 9 × 4 × 7 = 252
Probability of none of the bulbs being defective
⇒ 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
= 252 3,003
= 12 143
For Event B [Defectives] [Non-Defectives] Total Available 5 10 15 To Choose 1 4 5 Choices 5C1 10C4 15C5
Number of Favorable Choices
= The number of ways in which 1 defective and 5 non-defective
bulbs can be selected from the total 15 bulbs ⇒ mA = (Number of ways in which 1 defective bulbs can be selected from the 5)
× (Number of ways in which 4 non-defective bulbs can be selected
from the 10) 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.
= 5C1 × 10C4 = 5 × 10 × 9 × 8 × 7 4 × 3 × 2 × 1
= 5 × 10 × 3 × 7 = 1,050
Probability of only one of the bulbs being defective
⇒ Probability of occurrence of Event B
= Number of Favorable Choices for the Event Total Number of Possible Choices for the Experiment
⇒ P(B) = mB n
= 1,050 3,003
= 50 143
For Event C Probability of at least one of the bulbs being defective
= 1 − Probability of none of the bulbs being defective P(C) = 1 − P(A) = 1 − 12 143
= 143 − 12 143
= 131 143
For Event C [Alternative] Fundamental Counting Theorem (of Addition): Where an event can be accomplished in a number of alternative ways, the total number of ways in which the event can be accomplished is equal to the sum of the number of ways in which the alternative events can be accomplished.
Event C can be accomplished in two alternative ways.
- C1 : Selecting 1 defective and 4 non-defectives
- C2 : Selecting 2 defectives and 3 non-defectives
- C3 : Selecting 3 defectives and 2 non-defectives
- C4 : Selecting 4 defectives and 1 non-defectives
- C5 : Selecting 5 defectives and 0 non-defectives
[Defectives] [Non-Defectives] Total Available 5 10 15 To Choose 1 4 5 C1 Choices 5C1 10C4 15C5 To Choose 2 3 5 C2 Choices 5C2 10C3 15C5 To Choose 3 2 5 C3 Choices 5C3 10C2 15C5 To Choose 4 1 5 C3 Choices 5C4 10C1 15C5 To Choose 5 0 5 C3 Choices 5C5 10C0 15C5
Total number of Favorable/Favorable choices for Event C
⇒ mC = mC1 + mC2 + mC3 + mC4 + mC5 = (5C1 × 10C4) × (5C2 × 10C3) × (5C3 × 10C2) × (5C4 × 10C1) × (5C5 × 10C0)
= (5 × 10 × 9 × 8 × 7 4 × 3 × 2 × 1
) + ( 5 × 4 2 × 1
× 10 × 9 × 8 3 × 2 × 1
) + ( 5 × 4 × 3 3 × 2 × 1
× 10 × 9 2 × 1
)
+ ( 5 × 4 × 3 × 2 4 × 3 × 2 × 1
× 10) + (1)
= (1,050) + (1,200) + (450) + (50) + (1) = 2,751
Probability of at least one of the bulbs being defective
⇒ Probability of occurrence of Event C
= Number of Favorable Choices for the Event Total Number of Possible Choices for the Experiment
⇒ P(C) = mC n
= 2,751 3,003
= 131 143
| [Defectives] | [Non-Defectives] | Total | |
|---|---|---|---|
| Available | 5 | 10 | 15 |
| To Choose | 0 | 5 | 5 |
| Choices | 5C0 | 10C5 | 15C5 |
Number of Favorable Choices
| = | The number of ways in which 5 non-defective bulbs can be selected from the total 15 bulbs | |||
| ⇒ mA | = | (Number of ways in which 5 non-defective bulbs can be selected from the 10) | ||
|---|---|---|---|---|
| = | 10C5 | |||
| = |
| |||
| = | 9 × 4 × 7 | |||
| = | 252 |
Probability of none of the bulbs being defective
⇒ Probability of occurrence of Event A
| = |
| |||
| ⇒ P(A) | = |
| ||
|---|---|---|---|---|
| = |
| |||
| = |
|
| [Defectives] | [Non-Defectives] | Total | |
|---|---|---|---|
| Available | 5 | 10 | 15 |
| To Choose | 1 | 4 | 5 |
| Choices | 5C1 | 10C4 | 15C5 |
Number of Favorable Choices
| = | The number of ways in which 1 defective and 5 non-defective bulbs can be selected from the total 15 bulbs | |||||
| ⇒ mA | = | (Number of ways in which 1 defective bulbs can be selected from the 5) × (Number of ways in which 4 non-defective bulbs can be selected from the 10)
| ||||
|---|---|---|---|---|---|---|
| = | 5C1 × 10C4 | |||||
| = |
| |||||
| = | 5 × 10 × 3 × 7 | |||||
| = | 1,050 |
Probability of only one of the bulbs being defective
⇒ Probability of occurrence of Event B
| = |
| |||
| ⇒ P(B) | = |
| ||
|---|---|---|---|---|
| = |
| |||
| = |
|
For Event C Probability of at least one of the bulbs being defective
= 1 − Probability of none of the bulbs being defective P(C) = 1 − P(A) = 1 − 12 143
= 143 − 12 143
= 131 143
For Event C [Alternative] Fundamental Counting Theorem (of Addition): Where an event can be accomplished in a number of alternative ways, the total number of ways in which the event can be accomplished is equal to the sum of the number of ways in which the alternative events can be accomplished.
Event C can be accomplished in two alternative ways.
- C1 : Selecting 1 defective and 4 non-defectives
- C2 : Selecting 2 defectives and 3 non-defectives
- C3 : Selecting 3 defectives and 2 non-defectives
- C4 : Selecting 4 defectives and 1 non-defectives
- C5 : Selecting 5 defectives and 0 non-defectives
[Defectives] [Non-Defectives] Total Available 5 10 15 To Choose 1 4 5 C1 Choices 5C1 10C4 15C5 To Choose 2 3 5 C2 Choices 5C2 10C3 15C5 To Choose 3 2 5 C3 Choices 5C3 10C2 15C5 To Choose 4 1 5 C3 Choices 5C4 10C1 15C5 To Choose 5 0 5 C3 Choices 5C5 10C0 15C5
Total number of Favorable/Favorable choices for Event C
⇒ mC = mC1 + mC2 + mC3 + mC4 + mC5 = (5C1 × 10C4) × (5C2 × 10C3) × (5C3 × 10C2) × (5C4 × 10C1) × (5C5 × 10C0)
= (5 × 10 × 9 × 8 × 7 4 × 3 × 2 × 1
) + ( 5 × 4 2 × 1
× 10 × 9 × 8 3 × 2 × 1
) + ( 5 × 4 × 3 3 × 2 × 1
× 10 × 9 2 × 1
)
+ ( 5 × 4 × 3 × 2 4 × 3 × 2 × 1
× 10) + (1)
= (1,050) + (1,200) + (450) + (50) + (1) = 2,751
Probability of at least one of the bulbs being defective
⇒ Probability of occurrence of Event C
= Number of Favorable Choices for the Event Total Number of Possible Choices for the Experiment
⇒ P(C) = mC n
= 2,751 3,003
= 131 143
Probability of at least one of the bulbs being defective
| = | 1 − Probability of none of the bulbs being defective | |||||
| P(C) | = | 1 − P(A) | ||||
|---|---|---|---|---|---|---|
| = |
| |||||
| = |
| |||||
| = |
|
| Fundamental Counting Theorem (of Addition): Where an event can be accomplished in a number of alternative ways, the total number of ways in which the event can be accomplished is equal to the sum of the number of ways in which the alternative events can be accomplished. |
Event C can be accomplished in two alternative ways.
- C1 : Selecting 1 defective and 4 non-defectives
- C2 : Selecting 2 defectives and 3 non-defectives
- C3 : Selecting 3 defectives and 2 non-defectives
- C4 : Selecting 4 defectives and 1 non-defectives
- C5 : Selecting 5 defectives and 0 non-defectives
| [Defectives] | [Non-Defectives] | Total | ||
|---|---|---|---|---|
| Available | 5 | 10 | 15 | |
| To Choose | 1 | 4 | 5 | C1 |
| Choices | 5C1 | 10C4 | 15C5 | |
| To Choose | 2 | 3 | 5 | C2 |
| Choices | 5C2 | 10C3 | 15C5 | |
| To Choose | 3 | 2 | 5 | C3 |
| Choices | 5C3 | 10C2 | 15C5 | |
| To Choose | 4 | 1 | 5 | C3 |
| Choices | 5C4 | 10C1 | 15C5 | |
| To Choose | 5 | 0 | 5 | C3 |
| Choices | 5C5 | 10C0 | 15C5 |
Total number of Favorable/Favorable choices for Event C
| ⇒ mC | = | mC1 + mC2 + mC3 + mC4 + mC5 | ||||||||||||||||||||||||||
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
| = | (5C1 × 10C4) × (5C2 × 10C3) × (5C3 × 10C2) × (5C4 × 10C1) × (5C5 × 10C0) | |||||||||||||||||||||||||||
= |
| |||||||||||||||||||||||||||
| = | (1,050) + (1,200) + (450) + (50) + (1) | |||||||||||||||||||||||||||
| = | 2,751 |
Probability of at least one of the bulbs being defective
⇒ Probability of occurrence of Event C
| = |
| |||
| ⇒ P(C) | = |
| ||
|---|---|---|---|---|
| = |
| |||
| = |
|
Author : The Edifier