Drawing/Selecting/Picking Two/Three/More balls from a Box/Bag/Urn
Problem 5
Solution
Total number of balls in the bag
= | 3 Red + 7 Others |
= | 10 |
Number of balls drawn = 2
Experiment : Drawing 2 balls from the bag
Total Number of Possible Choices
= | Number of ways in which the two balls can be drawn from the total 12 | |||
⇒ n | = | 10C2 | ||
---|---|---|---|---|
= |
| |||
= | 45 |
Let
- A be the event of both the balls drawn being red
- B be the event that there will be at least one red ball
For Event A [Red] [Others] Total Available 3 7 10 To Choose 2 0 2 Choices 3C2 7C0 10C2
Number of Favorable Choices
= The number of ways in which the 2 red balls can be
drawn from the total 10 = 3C2 = 3 × 2 2 × 1
= 3
Probability of both the balls drawn being red
⇒ 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
= 3 45
= 1 15
• Odds
Number of Unfavorable Choices
= Total Number of possible choices − Number of Favorable choices ⇒ mAc = n − mA = 45 − 3 = 42
» in favor
Odds in Favor of both the balls drawn being red
⇒ Odds in Favor of Event A
= Number of Favorable Choices : Number of Unfavorable Choices = mA : mAc = 3 : 42 = 1 : 14
» against
Odds against both the balls drawn being red
⇒ Odds against Event A
= Number ofUnfavorable Choices : Number of Favorable Choices = mAc : mA = 3 : 45 = 1 : 15
For Event B 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 B can be accomplished in two alternative ways.
- B1 : Drawing 1 Red Ball and 1 Other Ball
- B2 : Drawing 2 Red Balls
[Red] [Others] Total Available 3 7 10 To Choose 1 1 2 B1 Choices 3C1 7C1 10C2 To Choose 2 0 2 B2 Choices 3C2 7C0 10C2
» "B1"
Number of Favorable Choices
= The number of ways in which a red ball and another ball
can be drawn from the total 10 mB1 = (Number of ways in which 1 red ball can be drawn from the total 3)
× (Number of ways in which 1 other ball can be drawn from the total 7) = 3C1 × 7C1 = 3 × 7 = 21
» "B2"
Number of Favorable Choices
= The number of ways in which 2 red balls can be drawn
from the total 10 mB2 = mA = 3
Total number of Favorable/Favorable choices for Event B
⇒ mB = mB1 + mB2 = 21 + 3 = 24
Probability of drawing the balls such that there will be at least one red ball
⇒ 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
= 24 45
= 8 15
• Odds
Number of Unfavorable Choices
= Total Number of possible choices − Number of Favorable choices ⇒ mBc = n − mB = 66 − 47 = 19
» in favor
Odds in Favor of drawing the balls such that they are of different colors/colours
⇒ Odds in Favor of Event B
= Number of Favorable Choices : Number of Unfavorable Choices = mA : mBc = 47 : 19
» against
Odds against drawing the balls such that they are of different colors/colours
⇒ Odds against Event A
= Number ofUnfavorable Choices : Number of Favorable Choices = mBc : mB = 19 : 47
[Red] | [Others] | Total | |
---|---|---|---|
Available | 3 | 7 | 10 |
To Choose | 2 | 0 | 2 |
Choices | 3C2 | 7C0 | 10C2 |
Number of Favorable Choices
= | The number of ways in which the 2 red balls can be drawn from the total 10 | ||
= | 3C2 | ||
= |
| ||
= | 3 |
Probability of both the balls drawn being red
⇒ 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 |
---|---|---|
= | 45 − 3 | |
= | 42 |
» in favor
Odds in Favor of both the balls drawn being red
⇒ Odds in Favor of Event A
= | Number of Favorable Choices : Number of Unfavorable Choices |
= | mA : mAc |
= | 3 : 42 |
= | 1 : 14 |
» against
Odds against both the balls drawn being red
⇒ Odds against Event A
= | Number ofUnfavorable Choices : Number of Favorable Choices |
= | mAc : mA |
= | 3 : 45 |
= | 1 : 15 |
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 B can be accomplished in two alternative ways.
- B1 : Drawing 1 Red Ball and 1 Other Ball
- B2 : Drawing 2 Red Balls
[Red] | [Others] | Total | ||
---|---|---|---|---|
Available | 3 | 7 | 10 | |
To Choose | 1 | 1 | 2 | B1 |
Choices | 3C1 | 7C1 | 10C2 | |
To Choose | 2 | 0 | 2 | B2 |
Choices | 3C2 | 7C0 | 10C2 |
» "B1"
Number of Favorable Choices
= | The number of ways in which a red ball and another ball can be drawn from the total 10 | |
mB1 | = | (Number of ways in which 1 red ball can be drawn from the total 3) × (Number of ways in which 1 other ball can be drawn from the total 7) |
---|---|---|
= | 3C1 × 7C1 | |
= | 3 × 7 | |
= | 21 |
» "B2"
Number of Favorable Choices
= | The number of ways in which 2 red balls can be drawn from the total 10 | |
mB2 | = | mA |
---|---|---|
= | 3 |
Total number of Favorable/Favorable choices for Event B
⇒ mB | = | mB1 + mB2 |
---|---|---|
= | 21 + 3 | |
= | 24 |
Probability of drawing the balls such that there will be at least one red ball
⇒ Probability of occurrence of Event B
= |
| |||
⇒ P(B) | = |
| ||
---|---|---|---|---|
= |
| |||
= |
|
• Odds
Number of Unfavorable Choices
= | Total Number of possible choices − Number of Favorable choices | |
⇒ mBc | = | n − mB |
---|---|---|
= | 66 − 47 | |
= | 19 |
» in favor
Odds in Favor of drawing the balls such that they are of different colors/colours
⇒ Odds in Favor of Event B
= | Number of Favorable Choices : Number of Unfavorable Choices |
= | mA : mBc |
= | 47 : 19 |
» against
Odds against drawing the balls such that they are of different colors/colours
⇒ Odds against Event A
= | Number ofUnfavorable Choices : Number of Favorable Choices |
= | mBc : mB |
= | 19 : 47 |