Probability for rolling three dice with the six sided dots such as 1, 2, 3, 4, 5 and 6 dots in each (three) dies. When three dice are thrown simultaneously/randomly, thus number of event can be 63 = (6 × 6 × 6) = 216 because each die has 1 to 6 number on its faces.Worked-out problems involving probability for rolling three dice: 1. Three dice are thrown together. Find the probability of: (i) getting a total of 5 (ii) getting a total of atmost 5 (iii) getting a total of at least 5. (iv) getting a total of 6. (v) getting a total of atmost 6. (vi) getting a total of at least 6. Solution: Three different dice are thrown at the same time. Therefore, total number of possible outcomes will be 63 = (6 × 6 × 6) = 216.(i) getting a total of 5: Number of events of getting a total of 5 = 6 i.e. (1, 1, 3), (1, 3, 1), (3, 1, 1), (2, 2, 1), (2, 1, 2) and (1, 2, 2) Therefore, probability of getting a total of 5 Number of favorable outcomesP(E1) = Total number of possible outcome = 6/216 = 1/36 (ii) getting a total of atmost 5: Number of events of getting a total of atmost 5 = 10 i.e. (1, 1, 1), (1, 1, 2), (1, 2, 1), (2, 1, 1), (1, 1, 3), (1, 3, 1), (3, 1, 1), (2, 2, 1) and (1, 2, 2). Therefore, probability of getting a total of atmost 5 Number of favorable outcomesP(E2) = Total number of possible outcome = 10/216 = 5/108 (iii) getting a total of at least 5: Number of events of getting a total of less than 5 = 4 i.e. (1, 1, 1), (1, 1, 2), (1, 2, 1) and (2, 1, 1). Therefore, probability of getting a total of less than 5 Number of favorable outcomesP(E3) = Total number of possible outcome = 4/216 = 1/54 Therefore, probability of getting a total of at least 5 = 1 - P(getting a total of less than 5) = 1 - 1/54 = (54 - 1)/54 = 53/54 (iv)
getting a total of 6: Number of events of getting a total of 6 = 10 i.e. (1, 1, 4), (1, 4, 1), (4, 1, 1), (1, 2, 3), (1, 3, 2), (2, 1, 3), (2, 3, 1), (3, 1, 2), (3, 2, 1) and (2, 2, 2). Therefore, probability of getting a total of 6 Number of favorable outcomesP(E4) = Total number of possible outcome = 10/216 = 5/108 (v) getting a total of atmost 6: Number of events of getting a total of atmost 6 = 20 i.e. (1, 1, 1), (1, 1, 2), (1, 2, 1), (2, 1, 1), (1, 1, 3), (1, 3, 1), (3, 1, 1), (2, 2, 1), (1, 2, 2), (1, 1, 4), (1, 4, 1), (4, 1, 1), (1, 2, 3), (1, 3, 2), (2, 1, 3), (2, 3, 1), (3, 1, 2), (3, 2, 1) and (2, 2, 2). Therefore, probability of getting a total of atmost 6 Number of favorable outcomesP(E5) = Total number of possible outcome = 20/216 = 5/54 (vi) getting a total of at least 6: Number of events of getting a total of less than 6 (event of getting a total of 3, 4 or 5) = 10 i.e. (1, 1, 1), (1, 1, 2), (1, 2, 1), (2, 1, 1) (1, 1, 3), (1, 3, 1), (3, 1, 1), (1, 2, 2), (2, 1, 2), (2, 2, 1). Therefore, probability of getting a total of less than 6 Number of favorable outcomesP(E6) = Total number of possible outcome = 10/216 = 5/108 Therefore, probability of getting a total of at least 6 = 1 - P(getting a total of less than 6) = 1 - 5/108 = (108 - 5)/108 = 103/108 These examples will help us to solve different types of problems based on probability for rolling three 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 Three 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.
Dice provide great illustrations for concepts in probability. The most commonly used dice are cubes with six sides. Here, we will see how to calculate probabilities for rolling three standard dice. It is a relatively standard problem to calculate the probability of the sum obtained by rolling two dice. There are a total of 36 different rolls with two dice, with any sum from 2 to 12 possible. How does the problem change if we add more dice? Just as one die has six outcomes and two dice have 62 = 36 outcomes, the probability experiment of rolling three dice has 63 = 216 outcomes. This idea generalizes further for more dice. If we roll n dice then there are 6n outcomes. We can also consider the possible sums from rolling several dice. The smallest possible sum occurs when all of the dice are the smallest, or one each. This gives a sum of three when we are rolling three dice. The greatest number on a die is six, which means that the greatest possible sum occurs when all three dice are sixes. The sum of this situation is 18. When n dice are rolled, the least possible sum is n and the greatest possible sum is 6n.
As discussed above, for three dice the possible sums include every number from three to 18. The probabilities can be calculated by using counting strategies and recognizing that we are looking for ways to partition a number into exactly three whole numbers. For example, the only way to obtain a sum of three is 3 = 1 + 1 + 1. Since each die is independent from the others, a sum such as four can be obtained in three different ways:
Further counting arguments can be used to find the number of ways of forming the other sums. The partitions for each sum follow:
When three different numbers form the partition, such as 7 = 1 + 2 + 4, there are 3! (3x2x1) different ways of permuting these numbers. So this would count toward three outcomes in the sample space. When two different numbers form the partition, then there are three different ways of permuting these numbers. We divide the total number of ways to obtain each sum by the total number of outcomes in the sample space, or 216. The results are:
As can be seen, the extreme values of 3 and 18 are least probable. The sums that are exactly in the middle are the most probable. This corresponds to what was observed when two dice were rolled.
|