Throwing/Tossing/Rolling Single/One Dice/Die - Probability - Problems & Solutions
Problem 1
(Or)
In a random experiment of rolling a die, what are the elementary events.
Solution
The events of the dice/die showing up the number
ONE TWO THREE FOUR FIVE SIX
⇒ The total number of possible choices in the experiment = 6
Problem 2
Solution
Total Number of Possible Choices
= 6 {ONE, TWO, THREE, FOUR, FIVE, SIX}
⇒ n = 6
Let
A : the event of getting an even integer.
For Event A
= 3 {TWO, FOUR, SIX}
⇒ mA = 3
Probability of getting an an even integer on rolling a dice
⇒ 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 |
= | 6 − 3 | |
= | 3 |
in favor
Odds in Favor of getting an even integer⇒ Odds in Favor of Event A
= Number of Favorable Choices : Number of Unfavorable Choices
= mA : mAc
= 3 : 3
= 1 : 1
against
Odds against getting an even integer⇒ Odds against Event A
= Number of Unfavorable Choices : Number of Favorable Choices
= mAc : mA
= 3 : 3
= 1 : 1
Problem 3
Solution
Total Number of Possible Choices
= 6 {ONE, TWO, THREE, FOUR, FIVE, SIX}
⇒ n = 6
Let
A : the event of getting an odd number or a multiple of 4.
For Event A
= 4 {ONE, THREE, FOUR, FIVE}
⇒ mA = 4
Probability of getting an odd number or a multiple of 4
⇒ 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 |
= | 6 − 4 | |
= | 2 |
in favor
Odds in Favor of getting an odd number or a multiple of 4⇒ Odds in Favor of Event A
= Number of Favorable Choices : Number of Unfavorable Choices
= mA : mAc
= 4 : 2
= 2 : 1
against
Odds against getting an odd number or a multiple of 4⇒ Odds against Event A
= Number of Unfavorable Choices : Number of Favorable Choices
= mAc : mA
= 2 : 4
= 1 : 2
Problem 4
Solution
Total Number of Possible Choices = 6 {ONE, TWO, THREE, FOUR, FIVE, SIX} ⇒ n = 6
Let A be the event of an even number more than 2 turning up.
For Event A Number of Favorable Choices
= 2 {FOUR, SIX} ⇒ mA = 2
Probability that an even number more than 2 turns up
⇒ 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
= 2 6
= 1 3
Probability that an even number more than 2 does not turn up
⇒ Probability of non-occurrence of Event A
= 1 − Probability of occurrence of Event A P(Ac) = 1 − P(A) = 1 − 1 3
= 3 − 1 3
= 2 3
• Alternative
Number of Unfavorable Choices
= Total Number of possible choices − Number of Favorable choices ⇒ mAc = n − mA = 6 − 2 = 4
Probability that an even number more than 2 does not turn up
⇒ Probability of non-occurrence of Event A
= Number of UnFavorable/Unfavorable Choices for the Event Total Number of Possible Choices for the Experiment
⇒ P(Ac) = mAc n
= 4 6
= 2 3
Number of Favorable Choices
= | 2 {FOUR, SIX} | |
⇒ mA | = | 2 |
---|
Probability that an even number more than 2 turns up
⇒ Probability of occurrence of Event A
= |
| |||
⇒ P(A) | = |
| ||
---|---|---|---|---|
= |
| |||
= |
|
Probability that an even number more than 2 does not turn up
⇒ Probability of non-occurrence of Event A
= | 1 − Probability of occurrence of Event A | |||||
P(Ac) | = | 1 − P(A) | ||||
---|---|---|---|---|---|---|
= |
| |||||
= |
| |||||
= |
|
• Alternative
Number of Unfavorable Choices
= | Total Number of possible choices − Number of Favorable choices | |
⇒ mAc | = | n − mA |
---|---|---|
= | 6 − 2 | |
= | 4 |
Probability that an even number more than 2 does not turn up
⇒ Probability of non-occurrence of Event A
= |
| |||
⇒ P(Ac) | = |
| ||
---|---|---|---|---|
= |
| |||
= |
|
Problem 5
Solution
Total Number of Possible Choices
= 6 {ONE, TWO, THREE, FOUR, FIVE, SIX}
⇒ n = 6
Let
A : the event of getting a number ≥ 4.
For Event A
= 3 {FOUR, FIVE, SIX}
⇒ mA = 3
Probability of getting a number ≥ 4
⇒ 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 |
= | 6 − 3 | |
= | 3 |
in favor
Odds in Favor of getting a number ≥ 4⇒ Odds in Favor of Event A
= Number of Favorable Choices : Number of Unfavorable Choices
= mA : mAc
= 3 : 3
= 1 : 1
against
Odds against getting a number ≥ 4⇒ Odds against Event A
= Number of Unfavorable Choices : Number of Favorable Choices
= mAc : mA
= 3 : 3
= 1 : 1
Problem 6
- Getting 4 when a dice is rolled
- Getting a face having a number less than 5?
- Throwing a number greater than 2.
- The number appearing on top is not an even number.
- Getting 3 and 5 simultaneously.
- Getting 4 or 6 in a throw of single die
- An an odd number less than 4 turns up
- An ace turns up
- Getting 7
- Getting an Even number or a multiple of 3
Solution
Total Number of Possible Choices
= 6 {ONE, TWO, THREE, FOUR, FIVE, SIX}
⇒ n = 6
Let
A : the event of getting 4 when the dice is rolled
For Event A
Number of Favorable Choices= 1 {FOUR}
⇒ mA = 1
Let
B : the event of getting a face having a number less than 5
For Event B
Number of Favorable Choices= 4 {ONE, TWO, THREE, FOUR}
⇒ mB = 4
Let
C : the event of throwing a number greater than 2
For Event C
Number of Favorable Choices= 4 {THREE, FOUR, FIVE, SIX}
⇒ mC = 4
Let
D : the event of the number appearing on top not being an even number
For Event D
Number of Favorable Choices= 3 {ONE, THREE, FIVE}
⇒ mD = 3
Let
E : the event of getting 3 and 5 simultaneously.
For Event E
Number of Favorable Choices= 0 {Φ}
⇒ mE = 0
Let
F : the event of getting 4 or 6 on a single throw.
For Event F
Number of Favorable Choices= 2 {FOUR, SIX}
⇒ mF = 2
Let
G : the event that an odd number less than 4 turns up.
For Event G
Number of Favorable Choices= 2 {ONE, THREE}
⇒ mG = 2
Let
H : the event that an ace turns up.
For Event H
Number of Favorable Choices= 1 {ONE}
⇒ mH = 1
Let
I : the event of getting 7.
For Event I
Number of Favorable Choices= 0 {Φ}
⇒ mI = 0
Since the die has only numbers from One to Six marked on it, the number 7 will not appear.
Let
J : the event of getting an Even number or a multiple of 3
For Event J
Number of Favorable Choices= 4 {TWO, THREE, FOUR, SIX}
⇒ mJ = 4
Practice Problems
- What is the chance of throwing a 5 with an ordinary dice?
- The probability of getting an odd number when we throw a single die is
- The probability of getting a number less than four when a die is rolled is __
- Find the probability of throwing a number greater than 4 when a die is rolled
- In a throw of a single die the probability of getting 3 or 5 is ___?
A dice is rolled, find
- P(even number)
- P(a number > 1)
- P(a number < 5)
- P(a number more than 6)
- P(a number < 7)
When a perfect die is rolled what is the probability of getting a face having
- 4 Points
- Odd Number
- 2 Points Or 3 Points
- Find the probability of getting 2 when a die is rolled
- What is the probability of throwing a number greater than 3 with an ordinary dice?
- A die is rolled. What is the probability that a number 1 or 6 may appear on the upper face?
- A (six-faced) die is thrown. Find the chance that any one of 1, 2, 3 turns up?
If a die is tossed, what is the probability that the number appearing on top is
- even
- less than 4
- not an even number
- either an even or an odd number
- an odd number less than 4.
- The probability of not getting 1, when a die is rolled
- If a die is tossed, what is the chance of getting an even number greater than 2
- Find the chance of not throwing an ace, two or three in a single throw with a die.
- what are the odds against throwing ace or six in a single throw with a die? And what are the odds in favour?