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Three dice are rolled. By assuming that each of the $6^3=216$possible outcomes is equally likely

In the random experiment of rolling three dice at a time, there are $6^3=216$potential results. Let X is the sum of the numerals upturned on the three dice. Then the potential values of X can vary from 3 Which is for the outcome (1,1,1) to 18 i.e. for the product (6,6,6).

Since all the 216 products are equally likely, as in the case of the two dice problem, it is also symmetric distribution. Outcomes 3 and 18 have the same probability, i.e. $\frac{1}{216}$.

Similarly the outcomes (4 and 17), (5 and 16), (6 and 15), (7 and 14), (8 and 13), (9 and 12) and (10 and 11) have the same probabilities, which we need to obtain now.

Probability of getting a sum of 4 with three dice is nothing but getting any of the outcomes (1,1, 2),(1,2,1), and (2,1,1). A total of 3 outcomes. Thus, the probability is $\frac{3}{216}$.

Probability of getting a sum of 5 with three dice is nothing but getting any of the outcomes (1,1,3), (1,3,1),(3,1,1),(1,2,2),(2,1,2), and (2,2,1). A total of 6 outcomes. Thus, the probability is $\frac{6}{216}$.

Probability of getting a sum of 6 with three dice is nothing but getting any of the outcomes (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). A total of 10 outcomes. So, the probability is $\frac{10}{216}$.

Probability of getting a sum of 7 with three dice is nothing but getting any of the outcomes (1,1,5),(1,5,1),(5,1,1),(1,2,4),(1,4,2),(2,1,4),(2,4,1),(4,1,2),(4,2,1),(1,3,3),(3,1,3),(3,3,1),(2,2,3),(2,3,2) and (3,2,2). A total of 15 outcomes. Thus, the probability is $\frac{15}{216}$.

Probability of getting a sum of 8 with three dice is nothing but getting any of the outcomes (1,1,6),(1,6,1),(6,1,1),(1,2,5),(1,5,2),(2,1,5),(2,5,1),(5,1,2),(5,2,1),(1,3,4),(1,4,3),(3, 1,4),(3,4,1),(4,1,3),(4,3,1),(2,2,4),(2,4,2),(4,2,2),(2,3,3),(3,2,3), and (3,3,2). A sum of 21 outcomes. Hence, the probability is $\frac{21}{216}$.

Probability of getting a sum of 9 with three dice is nothing but getting any of the outcomes (1,2,6),(1,6,2),(2,1,6),(2,6,1),(6,1,2),(6,2,1),(1,3,5),(1,5,3),(3,1,5),(3,5,1),(5,1,3),(5,3,1),(1,4,4),(4,1,4),(4,4,1),(2,2,5),(2,5,2),(5,2,2),(2,3,4),(2,4,3),(3,2,4),(3,4,2),(4,2,3),(4,3,2), and (3,3,3). A sum of 25 outcomes. Thus, the probability is $\frac{25}{216}$.

Probability of getting a sum of 10 with three dice is nothing but getting any of the outcomes (1, 3 ,6),(1,6,3),(3,1,6),(3,6,1),(6,1,3),(6,3,1),(1,4,5),(1,5,4),(4,1,5),(4,5,1),(5,1,4),(5,4,1),(2,2,6),(2,6,2),(6,2,2),(2,3,5),(2,5,3),(3,2,5),(3,5,2),(5,2,3),(5,3,2),(2,4,4),(4,2,4),(4,4,2),(3,3,4),(3,4,3), and (4,3,3). A sum of 27 outcomes. Thus, the probability is $\frac{27}{216}$.

The probability distribution of X can be tabulated as stated below: