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Difficulty:
75%
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Question Stats:
56% (02:27) correct
44%
(02:34)
wrong
based on 109
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If three cubical dice, each numbered from 1 to 6 inclusive, are rolled, and the facing numbers sum to 15, what is the probability that at least one of the facing numbers is a 6?
A) \( \frac{1}{2}\)
B) \(\frac{2}{3}\)
C) \(\frac{4}{5}\)
D) \(\frac{6}{7}\)
E) \(\frac{ 9}{10}\)
A) \( \frac{1}{2}\)
B) \(\frac{2}{3}\)
C) \(\frac{4}{5}\)
D) \(\frac{6}{7}\)
E) \(\frac{ 9}{10}\)
Hi,
For sum total of 15 - There are a total of following 10 possible scenarios
6,6,3 - 3 such cases
6,5,4 - 3! = 6 cases
5,5,5 - 1 such case
Hence, in 9 out of 10 cases - we get at least one 6. So, the answer should be 9/10, Option E
For sum total of 15 - There are a total of following 10 possible scenarios
6,6,3 - 3 such cases
6,5,4 - 3! = 6 cases
5,5,5 - 1 such case
Hence, in 9 out of 10 cases - we get at least one 6. So, the answer should be 9/10, Option E
