A standard six-sided die is rolled twice and the top faces are observed and the probability that the sum of the numbers on the top faces is 10 is $\frac{1}{12}$

<u>Solution: </u>

The sample space of two dice outcomes is given as follows:

{(1, 1) (1, 2) (1, 3) (1, 4) (1, 5) (1, 6)

(2, 1) (2,2) (2,3) (2,4) (2,5) (2,6)

(3, 1) (3,2) (3,3) (3,4) (3,5) (3,6)

(4, 1) (4,2) (4,3) (4,4) (4,5) (4,6)

(5, 1) (5,2) (5,3) (5,4) (5,5) (5,6)

(6, 1) (6,2) (6,3) (6,4) (6,5) (6,6) }

Total number of events n(s) = 36

Expected Outcomes = {(4, 6), (5, 5), (6, 4)}

Number of favorable outcomes n(E) = 3

Probability (sum of numbers on the top faces is 10) = $\frac{n(E)}{n(s)}$ = $\frac{3}{36}$ = $\frac{1}{12}$

Therefore the probability that the sum of the numbers on the top faces is 10 is $\frac{1}{12}$

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3 years ago