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

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

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