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

When two dice are thrown simultaneously, thus number of event can be 6Probability – 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:**

Therefore, probability of getting ‘six as a product’

P(E

= 4/36

= 1/9

**(ii) getting sum ≤**** 3:**

Therefore, probability of getting ‘sum ≤ 3’

P(E

= 3/36

= 1/12

**(iii) getting sum ≤**** 10:**

[(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’

P(E

= 33/36

= 11/12

Therefore, probability of getting ‘a doublet’

P(E

= 6/36

= 1/6

**(v)
getting a sum of 8:**

Therefore, probability of getting ‘a sum of 8’

P(E

= 5/36

**(vi)
getting sum divisible by 5:**

Therefore, probability of getting ‘sum divisible by 5’

P(E

= 7/36

**(vii)
getting sum of atleast 11:**

Therefore, probability of getting ‘sum of atleast 11’

P(E

= 3/36

= 1/12

**(viii) getting a
multiple of 3 as the sum:**

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

P(E

= 12/36

= 1/3

**(ix) getting a total
of atleast 10:**

Therefore, probability of getting ‘a total of atleast 10’

P(E

= 6/36

= 1/6

**(x) getting an even
number as the sum:**

Therefore, probability of getting ‘an even number as the sum

P(E

= 18/36

= 1/2

**(xi) getting a prime
number as the sum:**

Therefore, probability of getting ‘a prime number as the sum’

P(E

= 15/36

= 5/12

**(xii) getting a
doublet of even numbers:**

Therefore, probability of getting ‘a doublet of even numbers’

P(E

= 3/36

= 1/12

** **

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

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

P(E

= 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:**

E

⇒ P(E

Therefore, P(E

⇒ odds in favour of E

**(ii) the odds against getting the sum 6:**

E

⇒ P(E

Therefore, P(E

⇒ odds against E

**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)

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 of Tossing Two Coins**

**Probability of Tossing Three Coins**

**Probability for Rolling Two Dice**

**Probability for Rolling Three Dice**

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