Two different dice are thrown at the same time. Find the probability of getting : Show 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. understand what is the probability that the sum is it mean in the question we have to find the probability that two dice than the sum of the both died then restart the total outcome it can be first and second also album and it can 12 and it can we 1 and 31 and 41 and 5 and one and only 21 and 22 and 3 and 4 and 5 and 12 ncn 132 T3 T4 3 and 5 and 6 and it can be 41 42 43 44 45 and 46 and 51 52 53 54 55 and 56 2 3 4 5 and 6 are the possible outcome if we talk to diet and there are two possible outcome total possible outcome is equal to now we find the favourable out of it mean if the sum of the both 5 and here before 15 and 23 and 2 and 4 and 1 now it mean the four conditions are favorable total outcome and we know that probability to total outcome total possible outcome aur Jyada 36 will get it is equal to 1 then it mean the probability of two dice are tossed and the sum of the pipe and then discovered stage 1 by this your final answer thank you What is the probability of getting a sum of 5 or 8 when 2 dice are rolled once?
Answer (Detailed Solution Below)Option 2 : 1/4 Free Reading Comprehension Vol 1 12 Questions 36 Marks 20 Mins Formula Used: Probability = Possible outcomes/Total outcomes Calculations: When 2 dice are rolled, Total outcomes = 36 Possible outcomes of getting 5 = (1, 4), (4, 1), (2, 3), (3, 2) ⇒ Number of possible outcomes of getting 5 = 4 Possible outcomes of getting 8 = (2, 6), (6, 2), (3, 5), (5, 3), (4, 4) ⇒ Number of possible outcomes of getting 8 = 5 ⇒ Total number of possible outcomes = 4 + 5 = 9 Probability = Possible outcomes/Total outcomes ⇒ Probability = 9/36 = 1/4 ∴ The probability of getting a sum of 5 or 8 when 2 dice are rolled once is 1/4. Last updated on Sep 26, 2022 The National Testing Agency (NTA) made the Scoreboard Link Active for the CMAT (Common Management Admission Test) Exam 2022 on the 19th of July 2022. The candidates can check and download their CMAT Result by following the steps mentioned here. CMAT offers a gateway to get admission in a Post Graduation course in the top B-School of India. Candidates should continue their preparation as the notification for next year's examination is scheduled to be out very soon. 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 have (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. 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 =
4/36 (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 = 3/36 (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 = 33/36 (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 = 6/36 (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 = 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 = 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 = 3/36 (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
= 12/36 (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 = 6/36 (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 = 18/36 (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 = 15/36 (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 = 3/36 (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 = 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, (ii) the odds against getting the sum 6: Let E2 be the event of getting the sum 6. Then, 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 Random Experiments Experimental Probability Events in Probability Empirical Probability Coin Toss Probability Probability of Tossing Two Coins Probability of Tossing Three Coins Complimentary Events Mutually Exclusive Events Mutually Non-Exclusive Events Conditional Probability Theoretical Probability Odds and Probability Playing Cards Probability Probability and Playing Cards Probability for Rolling Two Dice Solved Probability Problems Probability for Rolling Three Dice 9th Grade Math From Probability for Rolling Two Dice to HOME PAGE Didn't find what you were looking for? Or want to know more information about Math Only Math. Use this Google Search to find what you need. What is the probability of rolling two dice and getting a sum of 5?The probability of getting a sum of 5 on rolling two dice is 1/11.
When two dice are thrown find the probability of getting product as 5?Hence, the probability of getting a product divisible by 5 is 3611.
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