A pair for fair dice is thrown. If the two numbers

14/15

Pembahasan

When two fair dice are thrown, the total number of possible outcomes is 6 * 6 = 36. Each outcome is an ordered pair (x, y), where x is the result of the first die and y is the result of the second die. We are given the condition that the two numbers appearing are different. Let A be the event that the two numbers appearing are different. The outcomes where the numbers are the same are (1,1), (2,2), (3,3), (4,4), (5,5), (6,6). There are 6 such outcomes. Therefore, the number of outcomes where the two numbers are different is 36 - 6 = 30. So, the size of our sample space is now 30. We want to find the probability that the sum is 4 or lots (meaning greater than or equal to 4). Let B be the event that the sum is 4 or greater than 4. We need to find the outcomes from the 30 possibilities (where the numbers are different) that satisfy this condition. It's easier to find the complement event: the sum is less than 4. The possible sums less than 4 are 2 and 3. The outcomes with a sum of 2 is (1,1). The outcomes with a sum of 3 are (1,2) and (2,1). We are given that the two numbers must be different. So, (1,1) is excluded. The outcomes with a sum of 3 are (1,2) and (2,1). Both have different numbers. So, there are 2 outcomes where the sum is less than 4 and the numbers are different. The probability of the sum being less than 4 (given numbers are different) is 2/30 = 1/15. The probability that the sum is 4 or greater than 4 (given numbers are different) is 1 - P(sum < 4 | numbers are different) = 1 - (1/15) = 14/15.