three dice are thrown together. find the probability of getting a sum of at least 6.
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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 outcomes
P(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
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 outcomes
P(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
Charu172:
thats what i said...
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