When two dices are thrown, there are (6 × 6) = 36 outcomes.The set of these outcomes is the sample space, which is given by S = (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)
Rolling two dice always plays a key role in probability concept. Whenever we go through the stuff probability in statistics, we will definitely have examples with rolling two dice. Sample Space When 2 Dice are RolledLook at the six faced die which is given below. The above six faced die has the numbers 1, 2, 3, 4, 5, 6 on its faces. When a die is rolled once, the sample space is S = {1, 2, 3, 4, 5, 6} So, total no. of all possible outcomes = 6 When two dice are rolled, total no. of all possible outcomes = 6 x 6 = 36 Here, the sample space is given when two dice are rolled Let us understand the sample space of rolling two dice. For example, (4, 3) stands for getting '4' on the first die and and '3' on the second die. (1, 6) stands for getting '1' on the first die and and '6' on the second die. Probability for Rolling 2 Dice - FormulaWe can use the formula from classic definition to find probability when two dice are rolled. or Since there are 36 outcomes in total when two dice are rolled, we have n(S) = 36 Solved ProblemsProblem 1 : A dice is rolled twice. What is the probability of getting a difference of 2 points? Solution : If an experiment results in p outcomes and if the experiment is repeated q times, then the total number of outcomes is pq. In the present case, since a dice results in 6 outcomes and the dice is rolled twice, total no. of outcomes or elementary events is 62 or 36. We assume that the dice is unbiased which ensures that all these 36 elementary events are equally likely. Now a difference of 2 points in the uppermost faces of the dice thrown twice can occur in the following cases : Thus denoting the event of getting a difference of 2 points by A, we find that the no. of outcomes favorable to A, from the above table, is 8. By classical definition of probability, we get P(A) = 8/36 P(A) = 2/9 Problem 2 : Two dice are thrown simultaneously. Find the probability that the sum of points on the two dice would be 7 or more. Solution : If two dice are thrown then, as explained in the last problem, total no. of elementary events is 62 or 36. Now a total of 7 or more i.e. 7 or 8 or 9 or 10 or 11 or 12 can occur only in the following combinations : Sum = 7 -----> (1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1) Sum = 8 -----> (2, 6), (3, 5), (4, 4), (5, 3), (6, 2) Sum = 9 -----> (3, 6), (4, 5), (5, 4), (6, 3) Sum = 10 -----> ((4, 6), (5, 5), (6, 4) Sum = 11 -----> (5, 6), (6, 5) Sum = 12 -----> (6, 6) Thus the no. of favorable outcomes is 21. Letting A stand for getting a total of 7 points or more, we have P(A) = 21/36 P(A) = 7/12 Problem 3 : Two dice are thrown simultaneously. Find the probability of getting a doublet. Solution : Let us look at the sample when two dice are rolled. In the sample space of rolling two dice, there are six cases when in which doubles are rolled. They are (1, 1), (2, 2), (3, 3), (4, 4), (5, 5) and (6, 6) Letting A stand for getting a doublet, we have P(A) = 6/36 P(A) = 1/6 Kindly mail your feedback to We always appreciate your feedback. ©All rights reserved. onlinemath4all.com
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 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): 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)] = 4Therefore, probability of getting ‘six as a product’ Number of favorable outcomesP(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)] = 3Therefore, probability of getting ‘sum ≤ 3’ Number of favorable outcomesP(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 outcomesP(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 outcomesP(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)] = 5Therefore, probability of getting ‘a sum of 8’ Number of favorable outcomesP(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)] = 7Therefore, probability of getting ‘sum divisible by 5’ Number of favorable outcomesP(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)] = 3Therefore, probability of getting ‘sum of atleast 11’ Number of favorable outcomesP(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)] = 12Therefore, probability of getting ‘a multiple of 3 as the sum’ Number of favorable outcomesP(E8) = Total number of possible outcome = 12/36 = 1/3 (ix) getting a total of atleast 10: Therefore, probability of getting ‘a total of atleast 10’ Number of favorable outcomesP(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)] = 18Therefore, probability of getting ‘an even number as the sum Number of favorable outcomesP(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)] = 15Therefore, probability of getting ‘a prime number as the sum’ Number of favorable outcomesP(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)] = 3Therefore, probability of getting ‘a doublet of even numbers’ Number of favorable outcomesP(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)] = 11Therefore, probability of getting ‘a multiple of 2 on one die and a multiple of 3 on the other die’ Number of favorable outcomesP(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) 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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