Question

A fair 6-sided die is rolled three times. Which is more likely: a sum of 11 or a sum of 12? Answer the question by calculating the probabilities for both. Thint 1] There are multiple ways to solve this problem. You may list all the favorable permutations to get the sum. However, this might be tedious and more error-prone. An easier way is to list only the favorable combinations (i.e., 3 numbers regardless of their order), and then find out the corresponding numbers of permutations for each combination. [hint 2] In a combination in which two numbers are the same (e.g., 2, 5, 5), there are 3 corresponding permutations (2-5-5, 5-2-5, and 5-5-2).

Answer 1

total possible outcomes for rolling a die three times=6³ = 216

favorable outcomes for sum of 11 are

(1,4,6) -->permutaion = 3! = 6

( 1,5,5) ---> permutaion = 3!/2! = 3

(2,3,6) -->permutaion=3! = 6

(2,4,5) -->permutaion = 3!=6

(3,4,4)-->permutaion=3!/2!=3

(3,5,3)-->permutaion=3

total permuation for summ of 11= 6+3+6+6+3+3+6 = 27

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favorable outcomes for sum of 12 are

(1,5,6)-->permutaion = 3!=6

(2,5,5)-->permutaion = 3

(2,4,6)-->permutaion = 3!=6

(3,4,5)-->permutaion = 3!=6

(3,3,6)-->permutaion = 3

(4,4,4)=1

total permuation for sum of 12 = 25

sp,P(11) = 27/216

P(12)=25/216

so, P(12) is greater than P(11)