To solve this problem, we need to determine the probability that the sum of the numbers rolled, $X$, is less than 7.

Step 1: Total possible outcomes

Two dice are rolled, and each die has 6 sides. Therefore, there are $6 \times 6 = 36$ equally likely outcomes.

Step 2: Calculate the outcomes where $X < 7$

The possible sums $X$ that are less than 7 are: $2, 3, 4, 5, 6$. We need to count how many outcomes correspond to these sums.

Case 1: $X = 2$

This happens when both dice show 1. So there is 1 outcome: $(1, 1)$.

Case 2: $X = 3$

This happens for: $(1, 2), (2, 1)$ So there are 2 outcomes.

Case 3: $X = 4$

This happens for: $(1, 3), (2, 2), (3, 1)$ So there are 3 outcomes.

Case 4: $X = 5$

This happens for: $(1, 4), (2, 3), (3, 2), (4, 1)$ So there are 4 outcomes.

Case 5: $X = 6$

This happens for: $(1, 5), (2, 4), (3, 3), (4, 2), (5, 1)$ So there are 5 outcomes.

Step 3: Total outcomes where $X < 7$

Adding up all these cases: $1 + 2 + 3 + 4 + 5 = 15$

Step 4: Probability calculation

The probability of $X < 7$ is the ratio of favorable outcomes to total outcomes: $P(X < 7) = \frac{\text{Number of outcomes where } X < 7}{\text{Total outcomes}} = \frac{15}{36} = \frac{5}{12}$

Final Answer:

$\boxed{P(X < 7) = \frac{5}{12}}$

Would you like a breakdown of other possible sums, or any additional explanations?

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Related Questions

1. How is the probability distribution of the sum $X$ constructed for two dice rolls?
2. What is the probability of $X = 7$?
3. How would you calculate $P(X > 8)$?
4. What is the expected value $E(X)$ of the sum of two dice rolls?
5. How does the symmetry of dice rolls affect the probability distribution?

Tip: When solving dice-related problems, organize outcomes systematically using tables or grids to ensure no cases are missed!