To solve the problem, we need to first determine the probability of rolling a sum of 5 and a sum of 7 with a pair of fair dice.
The possible combinations for each sum are as follows:
- For a sum of 5: (1,4), (2,3), (3,2), (4,1) — a total of 4 combinations.
- For a sum of 7: (1,6), (2,5), (3,4), (4,3), (5,2), (6,1) — a total of 6 combinations.
Now, we calculate the probabilities:
- Probability of rolling a sum of 5: P(5) = 4/36 = 1/9
- Probability of rolling a sum of 7: P(7) = 6/36 = 1/6
Next, we find the total probability of either getting a 5 or a 7:
P(5 or 7) = P(5) + P(7) = 1/9 + 1/6
Finding a common denominator (which is 18) gives us:
- P(5) = 2/18
- P(7) = 3/18
Thus:
P(5 or 7) = 2/18 + 3/18 = 5/18
Now, we want to find the conditional probability that a 5 occurs before a 7:
P(5 before 7) = P(5) / (P(5) + P(7))
Substituting in our values:
P(5 before 7) = (1/9) / ((1/9) + (1/6))
Since we already calculated the denominators as 5/18:
P(5 before 7) = (1/9) / (5/18) = (1/9) * (18/5) = 2/5
Therefore, the probability that a sum of 5 comes before a sum of 7 when rolling a pair of fair dice is 2/5.