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.Workedout 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
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