A dice is thrown twice. What is the probability of getting the sum as 11?

Concept: For an experiment with equally likely outcomes, the probability of an event is the number of outcomes favorable to that event divided by the total number of possible outcomes: P(Event) = Favorable outcomes / Total outcomes. When two dice are thrown, each die independently shows one of 6 faces, so the sample space contains 6 x 6 = 36 equally likely ordered outcomes.

Application:

1. Total outcomes: since each of the two throws has 6 possible results, the total sample space size is 6 x 6 = 36.

2. Favorable outcomes: list the ordered pairs (first throw, second throw) whose values add up to 11. Since each face is at most 6, both values must be close to 6 - the only such pairs are (5, 6) and (6, 5), giving 2 favorable outcomes.

3. Apply the formula: P(sum = 11) = 2/36.

4. Simplify: 2/36 = 1/18.

Cross-check: As an independent check, the sum distribution of two dice is symmetric about 7 - a sum of s occurs equally often as a sum of (14 minus s). Since 14 minus 11 = 3, and the outcomes summing to 3 are (1, 2) and (2, 1), i.e. 2 outcomes, the same count (2 out of 36) is confirmed for a sum of 11, matching the result 1/18.