Problem 5
Three electric lamps are fitted in a room. Three bulbs are chosen at random from 10 bulbs having 6 good bulbs . The chance that the room lighted is
Solution
Total number of bulbs in the lot
| = | 4 Defective + 6 Good |
| = | 10 |
Number of bulbs selected = 3
Experiment : Selecting 3 bulbs from the lot
Total Number of Possible Choices
| = | Number of ways in which the 3 bulbs can be selected from the total 10 bulbs |
| ⇒ n | = | 10C3 |
|---|
| = | |
| = | 120 |
Let A be the event of the room being lighted
If the room is to be lighted there should be at least one good bulb in the three bulbs drawn.
Thus Event A ⇒ The event of the bulbs drawn having at least one good bulb
For Event 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 A can be accomplished in three alternative ways.
- A1 : Selecting 2 defective and 1 good bulbs
- A2 : Selecting 1 defective and 2 good bulbs
- A3 : Selecting 0 defective and 3 good bulbs
| [Defectives] | [Good] | Total |
|---|
| Available | 4 | 6 | 10 |
|---|
| To Choose | 2 | 1 | 3 | A1 |
|---|
| Choices | 4C2 | 6C1 | 10C3 |
|---|
| To Choose | 1 | 2 | 3 | A2 |
|---|
| Choices | 4C1 | 6C2 | 10C3 |
|---|
| To Choose | 0 | 3 | 3 | A3 |
|---|
| Choices | 4C0 | 6C3 | 10C3 |
|---|
» "A1"
Number of Favorable Choices
| = | The number of ways in which the 2 defective and 1 good
bulbs can be selected from the total 10 bulbs |
| ⇒ mA | = | (Number of ways in which 2 defective bulbs can be selected from the 4)
× (Number of ways in which 1 good bulb can be selected from the 6) | 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. |
|
|---|
| = | 4C2 × 6C1 |
| = | |
| = | 36 |
» "A2"
Number of Favorable Choices
| = | The number of ways in which the 1 defective and 2 good
bulbs can be selected from the total 10 bulbs |
| ⇒ mA | = | (Number of ways in which 1 defective bulbs can be selected from the 4)
× (Number of ways in which 2 good bulb can be selected from the 6) |
|---|
| = | 4C1 × 6C2 |
| = | |
| = | 60 |
» "A3"
Number of Favorable Choices
| = | The number of ways in which the 3 good bulbs can be
selected from the total 10 bulbs |
| ⇒ mA | = | (Number of ways in which 3 good bulb can be selected from the 6) |
|---|
| = | 6C3 |
| = | |
| = | 20 |
Total number of Favorable/Favorable choices for Event A
| ⇒ mA | = | mA1 + mA2 + mA3 |
|---|
| = | 36 + 60 + 20 |
| = | 116 |
Probability of the room being lighted
⇒ Probability of occurrence of Event A
| = | | Number of Favorable Choices for the Event | | Total Number of Possible Choices for the Experiment |
|
| ⇒ P(A) | = | |
|---|
| = | |
| = | |
For Event A [Alternative] Probability of at least one of the bulbs being good
| = | 1 − Probability of none of the bulbs being good |
| P(A) | = | 1 − P(B) |
|---|
| = | |
| = | |
| = | |
Let B be the event of all the bulbs being defective
For Event B | [Defectives] | [Good] | Total |
|---|
| Available | 4 | 6 | 10 |
|---|
| To Choose | 3 | 0 | 3 |
|---|
| Choices | 4C3 | 6C0 | 10C3 |
|---|
Number of Favorable Choices
| = | The number of ways in which the 3 defective bulbs can
be selected from the total 4 bulbs |
| ⇒ mC | = | 4C3 |
|---|
| = | 4C(4 − 3)nCr = nC(n − r) |
| = | 4C1 |
| = | 4 |
Probability of all 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) | = | |
|---|
| = | |
| = | |
