Answer to Question #288327 in Statistics and Probability for ASIF MOHAMMED

Question

A six-sided die is rolled three times independently. Which is more likely: a sum of 11 or a sum of 12?

Answer 1

The total number of outcomes when a die is rolled 3 times is $6^3=216$ outcomes

A sum of 11 is obtained with the following six combinations.

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

A combination of $3$ distinct numbers corresponds to $^3P_1=3\times2=6$ permutations. Each combination of 3, two of which are equal corresponds to $^3P_2=3$ permutations

Therefore, the combinations $(6,4,1),(6,3,2),(5,4,2)$ have six permutations each while the combinations $(5,5,1),(5,3,3),(4,4,3)$ have three combinations each.

Thus, total number of combinations with sum equal to 11 is 6+6+6+3+3+3=27.

The probability of getting a combination with sum equal to 11 is ${27\over216}$.

A sum of 12 is obtained with the following 6 combinations.

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

A combination of $3$ distinct numbers corresponds to $^3P_1=3\times2=6$ permutations. Each combination of 3, two of which are equal corresponds to $^3P_2=3$ permutations and a combination of 3, three of which are equal corresponds to $^3P_0=1$ permutations.

Therefore, the combinations, $(6,4,2),(5,4,3),(6,5,1)$ have six combinations each. The combinations $(6,3,3),(5,5,2)$ have three combinations each and the combinations $(4,4,4)$ has one permutation.

Therefore, the total number of combinations with sum equal to 12 is 6+6+6+3+3+1=25. The probability of getting a combination with sum equal to 12 is ${25\over216}$.

Of the two events, the more likely event is the one with a higher probability of occurring. Therefore, the more likely event is a sum of 11 with a probability of $27\over216$ .