To solve the problem, let’s calculate the probability step by step:

Problem:

We roll two dice, each numbered from 1 to 6. We want to find the probability that the sum of the two dice is less than 5.

Step 1: Total Outcomes

Each die has 6 faces. When rolling two dice, the total number of outcomes is: $6 \times 6 = 36.$

Step 2: Favorable Outcomes

We need to find all outcomes where the sum of the two dice is less than 5. The possible sums are $2, 3,$ and $4$. Let's list the combinations for each sum:

• Sum = 2: (1, 1) → 1 outcome.
• Sum = 3: (1, 2), (2, 1) → 2 outcomes.
• Sum = 4: (1, 3), (2, 2), (3, 1) → 3 outcomes.

Thus, the total number of favorable outcomes is: $1 + 2 + 3 = 6.$

Step 3: Probability

The probability is the ratio of favorable outcomes to total outcomes: $P = \frac{\text{Favorable Outcomes}}{\text{Total Outcomes}} = \frac{6}{36} = \frac{1}{6}.$

Final Answer:

The probability that the sum of the two dice is less than 5 is: $\boxed{\frac{1}{6}}$

Would you like further details or explanations on this?

Related Questions:

1. What is the probability that the sum of the dice is greater than 9?
2. How many outcomes give a sum of exactly 7?
3. What is the probability of rolling a double (e.g., (1, 1), (2, 2), etc.)?
4. How can this calculation be extended to three dice?
5. How would the result change if the dice were biased?

Tip: