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Let's analyze the problem step by step:

We are working with two six-sided dice, one orange and one white, each of which has numbers from 1 to 6. The problem asks us to compute three probabilities based on the sum of the dice.

### 3.1.1 Probability that the sum is 12

The only way to get a sum of 12 is if both dice show a 6. There is exactly **1** favorable outcome: (6, 6). The total number of outcomes when rolling two dice is \( 6 \times 6 = 36 \).

Thus, the probability \( P(\text{sum} = 12) \) is:
\[
P(\text{sum} = 12) = \frac{1}{36}
\]

### 3.1.2 Probability that the sum is greater than or equal to 13

The maximum possible sum of two dice is 12 (which occurs with the combination (6, 6)), so there are **no outcomes** where the sum is 13 or greater.

Thus, the probability \( P(\text{sum} \geq 13) \) is:
\[
P(\text{sum} \geq 13) = 0
\]

### 3.1.3 Probability that the sum is less than 9

To solve this,

Two dice, one orange and one white, are thrown at the same time, and the numbers showing on the dice are added together. 3.1.1 Calculate the probability that the sum is 12. 3.1.2 Calculate the probability that the sum is greater than or equal to 13. 3.1.3 Calculate the probability that the sum is less than 9.

This page provides a detailed solution to a probability problem involving two six-sided dice. Learn how to calculate the probabilities of sums being exactly 12, greater than or equal to 13, and less than 9, with a step-by-step breakdown. Ideal for students in Grades 8-10.

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