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 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) = 4Therefore, 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) = 5Therefore, P(E2) = n(E2)/n(S) = 5/36 ⇒ odds against E2 = [1 – P(E2)]/P(E2) = (1 – 5/36)/(5/36) = 31/5.
These examples will help us to solve different types of problems based on probability for rolling two dice.
