What is the sample space of rolling a 6-sided die?

A fair die (fair dice) is a dice which isn't altered in any way, this means it has the same chance of landing on any of one of the faces. The opposite to this is a loaded dice or cheat dice.

1 Answer

Its {1,2,3,4,5,6}
which is actually a set of all the possible outcomes as the definition of sample space specifies.

When you roll a 6 sided dice, number of dots on uppermost face is called as outcome. Now, whenever a dice is rolled we can get either 1, 2,3,4,5 or 6 dots on the upper most face..that is now outcome.

• Dice definition, small cubes of plastic, ivory, bone, or wood, marked on each side with one to six spots, usually used in pairs in games of chance or in gambling.
• There are actually two ten sided dice in a standard 7-dice D&D set. The faces of one are numbered 0 to 9, and the faces of the other go from 00 to 90 by tens. To make a 'percentile' roll, you roll both die and add them together, with one exception: a double zero roll is 100.

So experiment here is 'Rolling a 6 faced dice' and list of possible outcomes is '{1,2,3,4,5,6}'.

Sample space by its definition is list of all the possible outcomes of an experiment.
So answer to your question is
S={1,2,3,4,5,6}

I hope its clear.

Related questions

Probability for rolling two dice with the six sided dotssuch as 1, 2, 3, 4, 5 and 6 dots in each die.

When two dice are thrown simultaneously, thus number of event can be 6² = 36 because each die has 1 to 6 number on its faces. Then the possible outcomes are shown in the below table.

Probability – Sample space for two dice (outcomes):

Note:

(i) The outcomes (1, 1), (2, 2), (3, 3), (4, 4), (5, 5) and (6, 6) are called doublets.

(ii) The pair (1, 2) and (2, 1) are different outcomes.

Worked-out problems involving probability for rolling two dice:

1. Two dice are rolled. Let A, B, C be the events of getting a sum of 2, a sum of 3 and a sum of 4 respectively. Then, show that

(i) A is a simple event

(ii) B and C are compound events

(iii) A and B are mutually exclusive

Solution:

Clearly, we have
A = {(1, 1)}, B = {(1, 2), (2, 1)} and C = {(1, 3), (3, 1), (2, 2)}.

(i) Since A consists of a single sample point, it is a simple event.

(ii) Since both B and C contain more than one sample point, each one of them is a compound event.

(iii) Since A ∩ B = ∅, A and B are mutually exclusive.

2. Two dice are rolled. A is the event that the sum of the numbers shown on the two dice is 5, and B is the event that at least one of the dice shows up a 3.
Are the two events (i) mutually exclusive, (ii) exhaustive? Give arguments in support of your answer.

Solution:

When two dice are rolled, we have n(S) = (6 × 6) = 36.

Now, A = {(1, 4), (2, 3), (4, 1), (3, 2)}, and

B = {(3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6), (1,3), (2, 3), (4, 3), (5, 3), (6, 3)}

(i) A ∩ B = {(2, 3), (3, 2)} ≠ ∅.

Hence, A and B are not mutually exclusive.

(ii) Also, A ∪ B ≠ S.

Therefore, A and B are not exhaustive events.

More examples related to the questions on the probabilities for throwing two dice.

3. Two dice are thrown simultaneously. Find the probability of:

(i) getting six as a product

(ii) getting sum ≤ 3

(iii) getting sum ≤ 10

(iv) getting a doublet

(v) getting a sum of 8

(vi) getting sum divisible by 5

(vii) getting sum of atleast 11

(viii) getting a multiple of 3 as the sum

(ix) getting a total of atleast 10

(x) getting an even number as the sum

(xi) getting a prime number as the sum

(xii) getting a doublet of even numbers

(xiii) getting a multiple of 2 on one die and a multiple of 3 on the other die

Solution:

Two different dice are thrown simultaneously being number 1, 2, 3, 4, 5 and 6 on their faces. We know that in a single thrown of two different dice, the total number of possible outcomes is (6 × 6) = 36.

(i) getting six as a product:

Let E₁ = event of getting six as a product. The number whose product is six will be E₁ = [(1, 6), (2, 3), (3, 2), (6, 1)] = 4

Therefore, probability ofgetting ‘six as a product’

Number of favorable outcomes
P(E₁) = Total number of possible outcome
= 4/36
= 1/9

(ii) getting sum ≤ 3:

Let E₂ = event of getting sum ≤ 3. The number whose sum ≤ 3 will be E₂ = [(1, 1), (1, 2), (2, 1)] = 3

Therefore, probability ofgetting ‘sum ≤ 3’

Number of favorable outcomes
P(E₂) = Total number of possible outcome
= 3/36
= 1/12

(iii) getting sum ≤ 10:

Let E₃ = event of getting sum ≤ 10. The number whose sum ≤ 10 will be E₃ =

[(1,1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6),

(2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2,6),

(3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3,6),

(4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4,6)

(5, 1), (5, 2), (5, 3), (5, 4), (5, 5),

(6, 1), (6, 2), (6, 3), (6, 4)] = 33

Therefore, probability ofgetting ‘sum ≤ 10’

Number of favorable outcomes
P(E₃) = Total number of possible outcome
= 33/36
= 11/12
(iv)getting a doublet:Let E₄ = event of getting a doublet. The number which doublet will be E₄ = [(1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6)] = 6

Therefore, probability ofgetting ‘a doublet’

Number of favorable outcomes
P(E₄) = Total number of possible outcome
= 6/36
= 1/6

(v)getting a sum of 8:

Let E₅ = event of getting a sum of 8. The number which is a sum of 8 will be E₅ = [(2, 6), (3, 5), (4, 4), (5, 3), (6, 2)] = 5

Therefore, probability ofgetting ‘a sum of 8’

Number of favorable outcomes
P(E₅) = Total number of possible outcome
= 5/36

(vi)getting sum divisible by 5:

Let E₆ = event of getting sum divisible by 5. The number whose sum divisible by 5 will be E₆ = [(1, 4), (2, 3), (3, 2), (4, 1), (4, 6), (5, 5), (6, 4)] = 7

Therefore, probability ofgetting ‘sum divisible by 5’

Number of favorable outcomes
P(E₆) = Total number of possible outcome
= 7/36

(vii)getting sum of atleast 11:

Let E₇ = event of getting sum of atleast 11. The events of the sum of atleast 11 will be E₇ = [(5, 6), (6, 5), (6, 6)] = 3

Therefore, probability ofgetting ‘sum of atleast 11’

Number of favorable outcomes
P(E₇) = Total number of possible outcome
= 3/36
= 1/12

(viii) getting amultiple of 3 as the sum:

Let E₈ = event of getting a multiple of 3 as the sum. The events of a multiple of 3 as the sum will be E₈ = [(1, 2), (1, 5), (2, 1), (2, 4), (3, 3), (3, 6), (4, 2), (4, 5), (5, 1), (5, 4), (6, 3) (6, 6)] = 12

Therefore, probability ofgetting ‘a multiple of 3 as the sum’

Number of favorable outcomes
P(E₈) = Total number of possible outcome
= 12/36
= 1/3

(ix) getting a totalof atleast 10:

Let E₉ = event of getting a total of atleast 10. The events of a total of atleast 10 will be E₉ = [(4, 6), (5, 5), (5, 6), (6, 4), (6, 5), (6, 6)] = 6

Therefore, probability ofgetting ‘a total of atleast 10’

Number of favorable outcomes
P(E₉) = Total number of possible outcome
= 6/36
= 1/6

(x) getting an evennumber as the sum:

Let E₁₀ = event of getting an even number as the sum. The events of an even number as the sum will be E₁₀ = [(1, 1), (1, 3), (1, 5), (2, 2), (2, 4), (2, 6), (3, 3), (3, 1), (3, 5), (4, 4), (4, 2), (4, 6), (5, 1), (5, 3), (5, 5), (6, 2), (6, 4), (6, 6)] = 18

Therefore, probability ofgetting ‘an even number as the sum

Number of favorable outcomes
P(E₁₀) = Total number of possible outcome
= 18/36
= 1/2

(xi) getting a primenumber as the sum:

Let E₁₁ = event of getting a prime number as the sum. The events of a prime number as the sum will be E₁₁ = [(1, 1), (1, 2), (1, 4), (1, 6), (2, 1), (2, 3), (2, 5), (3, 2), (3, 4), (4, 1), (4, 3), (5, 2), (5, 6), (6, 1), (6, 5)] = 15

Therefore, probability ofgetting ‘a prime number as the sum’

Number of favorable outcomes
P(E₁₁) = Total number of possible outcome
= 15/36
= 5/12

(xii) getting adoublet of even numbers:

Let E₁₂ = event of getting a doublet of even numbers. The events of a doublet of even numbers will be E₁₂ = [(2, 2), (4, 4), (6, 6)] = 3

Therefore, probability ofgetting ‘a doublet of even numbers’

Number of favorable outcomes
P(E₁₂) = Total number of possible outcome
= 3/36
= 1/12

(xiii) getting amultiple of 2 on one die and a multiple of 3 on the other die:

Let E₁₃ = event of getting a multiple of 2 on one die and a multiple of 3 on the other die. The events of a multiple of 2 on one die and a multiple of 3 on the other die will be E₁₃ = [(2, 3), (2, 6), (3, 2), (3, 4), (3, 6), (4, 3), (4, 6), (6, 2), (6, 3), (6, 4), (6, 6)] = 11

Therefore, probability ofgetting ‘a multiple of 2 on one die and a multiple of 3 on the other die’

Number of favorable outcomes
P(E₁₃) = Total number of possible outcome
= 11/36

4. Twodice are thrown. Find (i) the odds in favour of getting the sum 5, and (ii) theodds against getting the sum 6.

Solution:

We know that in a single thrown of two die, the total numberof possible outcomes is (6 × 6) = 36.

Let S be the sample space. Then,n(S) = 36.

(i) the odds in favour of getting the sum 5:

Let E₁ be the event of getting the sum 5. Then,
E₁ = {(1, 4), (2, 3), (3, 2), (4, 1)}
⇒ P(E

Dice Singular

₁) = 4
Therefore, P(E₁) = n(E₁)/n(S) = 4/36 = 1/9
⇒ odds in favour of E₁ = P(E₁)/[1 – P(E₁)] = (1/9)/(1 – 1/9) = 1/8.

(ii) the odds against getting the sum 6:

Let E₂ be the event of getting the sum 6. Then,
E₂ = {(1, 5), (2, 4), (3, 3), (4, 2), (5, 1)}
⇒ P(E₂) = 5
Therefore, P(E₂) = n(E₂)/n(S) = 5/36
⇒ odds against E₂ = [1 – P(E₂)]/P(E₂) = (1 – 5/36)/(5/36) = 31/5.

5. Two dice, one blue and one orange, are rolled simultaneously. Find the probability of getting

(i) equal numbers on both

(ii) two numbers appearing on them whose sum is 9.

Solution:

The possible outcomes are

(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6),

(2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6),

(3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6),

(4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6)

(5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6)

(6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)

Dice Die

Therefore, total number of possible outcomes = 36.

What Is A Single Dice Called As A

(i) Number of favourable outcomes for the event E

= number of outcomes having equal numbers on both dice

= 6 [namely, (1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6)].

So, by definition, P(E) = (frac{6}{36})

= (frac{1}{6})

(ii) Number of favourable outcomes for the event F

= Number of outcomes in which two numbers appearing on them have the sum 9

= 4 [namely, (3, 6), (4, 5), (5, 4), (3, 6)].

Thus, by definition, P(F) = (frac{4}{36})

= (frac{1}{9}).

These examples will helpus to solve different types of problems based on probability for rollingtwo dice.

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