Practice GMAT Problem Solving Question

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A fair sided die labeled 1 to 6 is tossed three times. What is the probability the sum of the 3 throws is 16?
Correct Answer: C
  1. Probability problems require the knowledge of all possible combinations (mathematicians call this the probability space), the total number of combinations that satisfy the condition you are looking for and the probability that the combination will occur. The die is fair all possibilities are equally likely. Let A, B, C represent all possible values for the 3 die tosses. A = {1,2,3,4,5,6} , B = {1,2,3,4,5,6}, C = {1,2,3,4,5,6} and a,b,c represent possible outcomes of A,B,C respectively. (mathematicians say a,b,c are elements of the set A,B,C). We are to solve for the sum of 3 die tosses to equal 16. In other words, we are looking for all possible combinations such that a + b + c = 16.
  2. Each die toss has six possible outcomes. Tossing the die once gives six possible values. Tossing the die twice gives a total of ( 6 possibilities for the first toss) times ( 6 possibilities for the second toss) = 36 possible outcomes . Similarly, for three die tosses one has (6)(6)(6) = 216 possible outcomes.
  3. The first die toss is from 1 to 6. Since the maximum sum of the final 2 die tosses is 6 + 6 = 12, the first dice toss must be at least 4 for the sum of all three tosses to be 16.
  4. If the first die toss is 4 then the last two must add up to 12. This will occur for the case {4,6,6} which is a total of one possibility.
  5. If the first die toss is 5 then the last two must add up to 11. This will occur for the cases {5,5,6}, {5,6,5} which is a total of two possibilities.
  6. If the first die toss is 6 then the last two must add up to 10. This will occur for the case {6,4,6}, {6,5,5}, {6,6,4} which is a total of three possibilities.
  7. Out of the 216 total possible combinations, 1+2+3=6 add up to 6.
  8. The answer is given by (number of outcomes that add to 16 ) divided by (the total number of possible outcomes) which is 6 / 216 or 1 / 36

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