What is the probability of getting a sum of 2 when two dice are thrown?

Solution:

When two dice are thrown simultaneously, the sample space of the experiment is

{(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)}

So there are 36 equally likely outcomes.

Possible number of outcomes = 36.

(i)Let E be an event of getting a doublet.

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

Number of favourable outcomes = 6

P(E) = 6/36 = 1/6

Probability of getting a doublet is 1/6 .

(ii)Let E be an event of getting a sum of 8.

Favourable outcomes = {(2,6), (3,5), (4,4), (5,3), (6,2)}

Number of favourable outcomes = 5

P(E) = 5/36

Probability of getting a sum of 8 is 5/36.

One popular way to study probability is to roll dice. A standard die has six sides printed with little dots numbering 1, 2, 3, 4, 5, and 6. If the die is fair (and we will assume that all of them are), then each of these outcomes is equally likely. Since there are six possible outcomes, the probability of obtaining any side of the die is 1/6. The probability of rolling a 1 is 1/6, the probability of rolling a 2 is 1/6, and so on. But what happens if we add another die? What are the probabilities for rolling two dice?

To correctly determine the probability of a dice roll, we need to know two things:

  • The size of the sample space or the set of total possible outcomes
  • How often an event occurs

In probability, an event is a certain subset of the sample space. For example, when only one die is rolled, as in the example above, the sample space is equal to all of the values on the die, or the set (1, 2, 3, 4, 5, 6). Since the die is fair, each number in the set occurs only once. In other words, the frequency of each number is 1. To determine the probability of rolling any one of the numbers on the die, we divide the event frequency (1) by the size of the sample space (6), resulting in a probability of 1/6.

Rolling two fair dice more than doubles the difficulty of calculating probabilities. This is because rolling one die is independent of rolling a second one. One roll has no effect on the other. When dealing with independent events we use the multiplication rule. The use of a tree diagram demonstrates that there are 6 x 6 = 36 possible outcomes from rolling two dice.

Suppose that the first die we roll comes up as a 1. The other die roll could be a 1, 2, 3, 4, 5, or 6. Now suppose that the first die is a 2. The other die roll again could be a 1, 2, 3, 4, 5, or 6. We have already found 12 potential outcomes, and have yet to exhaust all of the possibilities of the first die.

The possible outcomes of rolling two dice are represented in the table below. Note that the number of total possible outcomes is equal to the sample space of the first die (6) multiplied by the sample space of the second die (6), which is 36.

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

The same principle applies if we are working on problems involving three dice. We multiply and see that there are 6 x 6 x 6 = 216 possible outcomes. As it gets cumbersome to write the repeated multiplication, we can use exponents to simplify work. For two dice, there are 62 possible outcomes. For three dice, there are 63 possible outcomes. In general, if we roll n dice, then there are a total of 6n possible outcomes.

With this knowledge, we can solve all sorts of probability problems:

1. Two six-sided dice are rolled. What is the probability that the sum of the two dice is seven?

The easiest way to solve this problem is to consult the table above. You will notice that in each row there is one dice roll where the sum of the two dice is equal to seven. Since there are six rows, there are six possible outcomes where the sum of the two dice is equal to seven. The number of total possible outcomes remains 36. Again, we find the probability by dividing the event frequency (6) by the size of the sample space (36), resulting in a probability of 1/6.

2. Two six-sided dice are rolled. What is the probability that the sum of the two dice is three?

In the previous problem, you may have noticed that the cells where the sum of the two dice is equal to seven form a diagonal. The same is true here, except in this case there are only two cells where the sum of the dice is three. That is because there are only two ways to get this outcome. You must roll a 1 and a 2 or you must roll a 2 and a 1. The combinations for rolling a sum of seven are much greater (1 and 6, 2 and 5, 3 and 4, and so on). To find the probability that the sum of the two dice is three, we can divide the event frequency (2) by the size of the sample space (36), resulting in a probability of 1/18.

3. Two six-sided dice are rolled. What is the probability that the numbers on the dice are different?

Again, we can easily solve this problem by consulting the table above. You will notice that the cells where the numbers on the dice are the same form a diagonal. There are only six of them, and once we cross them out we have the remaining cells in which the numbers on the dice are different. We can take the number of combinations (30) and divide it by the size of the sample space (36), resulting in a probability of 5/6.

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

What is the probability of getting a sum of 2 when two dice are thrown?

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):

What is the probability of getting a sum of 2 when two dice are thrown?

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 haveA = {(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)

What is the probability of getting a sum of 2 when two dice are thrown?

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.

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