Probability for Rolling Two Dice

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

Probability of Rolling Two Dice
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):

Probability for Rolling Two Dice

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 of getting ‘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 of getting ‘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 of getting ‘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 of getting ‘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 of getting ‘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 of getting ‘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 of getting ‘sum of atleast 11’

               Number of favorable outcomes
P(E7) =     Total number of possible outcome


      = 3/36
      = 1/12

(viii) getting a multiple 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 of getting ‘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 total of 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 of getting ‘a total of atleast 10’

               Number of favorable outcomes
P(E9) =     Total number of possible outcome


      = 6/36
      = 1/6

(x) getting an even number 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 of getting ‘an even number as the sum

               Number of favorable outcomes
P(E10) =     Total number of possible outcome


      = 18/36
      = 1/2

(xi) getting a prime number 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 of getting ‘a prime number as the sum’

               Number of favorable outcomes
P(E11) =     Total number of possible outcome


      = 15/36
      = 5/12

(xii) getting a doublet 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 of getting ‘a doublet of even numbers’

               Number of favorable outcomes
P(E12) =     Total number of possible outcome


      = 3/36
      = 1/12


(xiii) getting a multiple 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 of getting ‘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. Two dice are thrown. Find (i) the odds in favour of getting the sum 5, and (ii) the odds against getting the sum 6.

Solution:

We know that in a single thrown of two die, the total number of 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(E1) = 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)

Sample Space of Rolling a Pair of Dice

Therefore, total number of possible outcomes = 36.

(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 help us to solve different types of problems based on probability for rolling two dice.

Probability

Probability

Random Experiments

Experimental Probability

Events in Probability

Empirical Probability

Coin Toss Probability

Probability of Tossing Two Coins

Probability of Tossing Three Coins

Complimentary Events

Mutually Exclusive Events

Mutually Non-Exclusive Events

Conditional Probability

Theoretical Probability

Odds and Probability

Playing Cards Probability

Probability and Playing Cards

Probability for Rolling Two Dice

Solved Probability Problems

Probability for Rolling Three Dice










9th Grade Math

From Probability for Rolling Two Dice to HOME PAGE




Didn't find what you were looking for? Or want to know more information about Math Only Math. Use this Google Search to find what you need.



New! Comments

Have your say about what you just read! Leave me a comment in the box below. Ask a Question or Answer a Question.




Share this page: What’s this?

Recent Articles

  1. 2nd grade math Worksheets | Free Math Worksheets | By Grade and Topic

    Dec 04, 24 01:30 AM

    2nd Grade Math Worksheet
    2nd grade math worksheets is carefully planned and thoughtfully presented on mathematics for the students.

    Read More

  2. Time Duration |How to Calculate the Time Duration (in Hours & Minutes)

    Dec 04, 24 01:07 AM

    Time Duration Example
    Time duration tells us how long it takes for an activity to complete. We will learn how to calculate the time duration in minutes and in hours. Time Duration (in minutes) Ron and Clara play badminton…

    Read More

  3. Worksheet on Subtraction of Money | Real-life Word Problems | Answers

    Dec 04, 24 12:45 AM

    Worksheet on Subtraction of Money
    Practice the questions given in the worksheet on subtraction of money by using without conversion and by conversion method (without regrouping and with regrouping). Note: Arrange the amount of rupees…

    Read More

  4. Worksheet on Addition of Money | Questions on Adding Amount of Money

    Dec 04, 24 12:06 AM

    Worksheet on Addition of Money
    Practice the questions given in the worksheet on addition of money by using without conversion and by conversion method (without regrouping and with regrouping). Note: Arrange the amount of money in t…

    Read More

  5. Worksheet on Money | Conversion of Money from Rupees to Paisa

    Dec 03, 24 11:37 PM

    Worksheet on Money
    Practice the questions given in the worksheet on money. This sheet provides different types of questions where students need to express the amount of money in short form and long form

    Read More