Question: An experiment consists of rolling two fair dice and adding the dots on the two sides facing up…

An experiment consists of rolling two fair dice and adding the dots on the two sides facing up. Find the probability of the sum of the dots indicated. A sum less than or equal to 10 … The probability of getting a sum less than or equal to 10 is \boxed{}. (Type an integer or a simplified fraction.)

Solution

To solve this problem, we need to calculate the probability of getting a sum of 10 or less when rolling two six-sided dice. First, determine the total number of possible outcomes when rolling two dice. Each die has 6 faces, so: \[ 6 \times 6 = 36 \] Next, count the number of outcomes where the sum is 10 or less. We can list these outcomes by considering the sum of the two dice: - Sum = 2: (1, 1) - Sum = 3: (1, 2), (2, 1) - Sum = 4: (1, 3), (2, 2), (3, 1) - Sum = 5: (1, 4), (2, 3), (3, 2), (4, 1) - Sum = 6: (1, 5), (2, 4), (3, 3), (4, 2), (5, 1) - Sum = 7: (1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1) - Sum = 8: (2, 6), (3, 5), (4, 4), (5, 3), (6, 2) - Sum = 9: (3, 6), (4, 5), (5, 4), (6, 3) - Sum = 10: (4, 6), (5, 5), (6, 4) Count these outcomes: \[ 1 + 2 + 3 + 4 + 5 + 6 + 5 + 4 + 3 = 33 \] The probability is the number of favorable outcomes divided by the total number of possible outcomes: \[ \frac{33}{36} \] Simplify the fraction: \[ \frac{11}{12} \] Thus, the probability of getting a sum less than or equal to 10 is \(\frac{11}{12}\).