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 6^{3} = (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 6^{3} = (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 outcomes(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 outcomes(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 outcomesTherefore, 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 outcomes(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 outcomes(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 outcomesTherefore, 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 of Tossing Two Coins
Probability of Tossing Three Coins
Probability for Rolling Two Dice
Probability for Rolling Three Dice
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.
Nov 07, 24 12:38 AM
Nov 06, 24 11:59 PM
Nov 05, 24 01:49 PM
Nov 05, 24 01:15 AM
Nov 05, 24 12:55 AM
New! Comments
Have your say about what you just read! Leave me a comment in the box below. Ask a Question or Answer a Question.