After rolling a pair of fair, six-sided dice, what is the probability that the sum of the outcomes is at least 4 but less than 12?

(Round your answer to the third decimal place)

The two dice are assumed distinct.

How many ways are there to get the desired outcomes (success):
Sum
4 1-3,2-2,3-1 so 3
5 1-4,2-3,3,2,4,1 so 4
6 1-5,2-4,3-3,4-2,5-1 so 5
7 1-6,2-5,3-4,4-3,5-2,6-1 so 6
...
Continue until 11 (less than 12) and add the total number of ways to get success.
The number of possible outcomes (Ω) is 6×6=36.
Divide the number of successful outcomes by the number of possible outcomes to get the probability of success.

To find the probability of the sum of the outcomes being at least 4 but less than 12, we need to count the number of favorable outcomes and divide it by the total number of possible outcomes.

First, let's determine the favorable outcomes. The sum of the outcomes can be between 4 and 11 (inclusive). We can count these outcomes by listing all the possible pairs of numbers that satisfy this condition:

(1, 3), (1, 4), (1, 5), (1, 6)
(2, 2), (2, 3), (2, 4), (2, 5), (2, 6)
(3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6)
(4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6)
(5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6)
(6, 1), (6, 2), (6, 3), (6, 4), (6, 5)

Counting these outcomes, we have a total of 30 pairs.

Next, let's determine the total number of possible outcomes. Since we are rolling a pair of six-sided dice, each die has 6 possible outcomes. So, the total number of possible outcomes is 6 * 6 = 36.

To find the probability, we divide the number of favorable outcomes by the total number of possible outcomes:

Probability = Favorable outcomes / Total outcomes = 30 / 36 = 5/6 ≈ 0.833

Therefore, the probability that the sum of the outcomes is at least 4 but less than 12 is approximately 0.833, rounded to the third decimal place.