# GMAT Challenge Series (700+): Q1

Question:

What is the probability of getting a sum of 8 or 14 when rolling 3 dice simultaneously?

A. 1/6

B. 1/4

C. 1/2

D. 21/216

E. 32/216

$$\text{Possible options to get 8: 21 ways}$$ $$\text{(6,1,1): 3 ways}$$ $$\text{(5,2,1): 6 ways}$$ $$\text{(4,3,1): 6 ways}$$ $$\text{(4,2,2): 3 ways}$$ $$\text{(3,3,2): 3 ways}$$  $$\text{Possible options to get 14: 15 ways}$$ $$\text{(6,4,4): 3 ways}$$ $$\text{(6,5,3): 6 ways}$$ $$\text{(6,6,2): 3 ways}$$ $$\text{(5,5,4): 3 ways}$$  $$\text{Possible ways to get 8 or 14} = {15} + {21} = {36}$$ $$\text{Total combinations} = {6} \times {6} \times {6} = {216}$$  $$\text{Probability (Sum of 8 or 14 with 3 dice):} = {\text{Possible ways to get 8 or 14} \over \text{Total combinations}}$$ $$\text{Probability (Sum of 8 or 14 with 3 dice):} = \frac{36}{216} = \frac{1}{6}$$
$$\text{Probabilities of sums with 1 die:}$$ $${1} = \frac{1}{6}$$ $${2} = \frac{1}{6}$$ $${3} = \frac{1}{6}$$ $${4} = \frac{1}{6}$$ $${5} = \frac{1}{6}$$ $${6} = \frac{1}{6}$$  $$\text{Probabilities of sums with 2 dice:}$$ $${2} = \frac{1}{36}$$ $${3} = \frac{2}{36}$$ $${4} = \frac{3}{36}$$ $${5} = \frac{4}{36}$$ $${6} = \frac{5}{36}$$ $${7} = \frac{6}{36}$$ $${8} = \frac{5}{36}$$ $${9} = \frac{4}{36}$$ $${10} = \frac{3}{36}$$ $${11} = \frac{2}{36}$$ $${12} = \frac{1}{36}$$  $$\text{Probability of 8 with 3 dice:}$$ $$\text{Roll a 1 and then 7 on the next two dice}$$ $$\text{Roll a 2 and then 6 on the next two dice}$$ $$\text{...}$$ $$\text{Roll a 6 and then 2 on the next two dice}$$ $$\text{Probability (Sum of 8 with 3 dice):} = \frac{1}{6} \left({{6} + {5} + {4} + {3} + {2} + {1}} \over {36}\right) = \frac{1}{6} \left({21} \over {36}\right) = \frac{21}{216}$$  $$\text{Probability of 14 with 3 dice:}$$ $$\text{Roll a 2 and then 12 on the next two dice}$$ $$\text{Roll a 3 and then 11 on the next two dice}$$ $$\text{...}$$ $$\text{Roll a 6 and then 8 on the next two dice}$$ $$\text{Probability (Sum of 14 with 3 dice):} = \frac{1}{6} \left({{5} + {4} + {3} + {2} + {1}} \over {36}\right) = \frac{1}{6} \left({15} \over {36}\right) = \frac{15}{216}$$  $$\text{Probability (Sum of 8 or 14 with 3 dice):} = \frac{21}{216} + \frac{15}{216} = \frac{36}{216} = \frac{1}{6}$$