Question

If two fair dice are tossed, what is the probability that the sum is \(i, i=2,3, \ldots, 12 ?\)

Short answer

The probabilities of obtaining each sum (i=2 to 12) when two fair dice are tossed are: P(sum=2) = \(\frac{1}{36}\), P(sum=3) = \(\frac{1}{18}\), P(sum=4) = \(\frac{1}{12}\), P(sum=5) = \(\frac{1}{9}\), P(sum=6) = \(\frac{5}{36}\), P(sum=7) = \(\frac{1}{6}\), P(sum=8) = \(\frac{5}{36}\), P(sum=9) = \(\frac{1}{9}\), P(sum=10) = \(\frac{1}{12}\), P(sum=11) = \(\frac{1}{18}\), P(sum=12) = \(\frac{1}{36}\).

Step-by-step solution

List out all possible outcomes

List all the possible outcomes of throwing two dice - Each die can show a number from 1 to 6, making 36 possible combinations in total. We can represent the outcomes as an ordered pair (a, b) where 'a' denotes the number on the first dice and 'b' denotes the number on the second dice.

Identify the favorable outcomes for each sum from 2 to 12

For each value i (from 2 to 12), we need to count how many outcomes give us that sum. This can be done using the following process: 1. For sum = 2: There is 1 pair, (1,1). 2. For sum = 3: There are 2 pairs, (1,2) and (2,1). 3. For sum = 4: There are 3 pairs, (1,3), (2,2), and (3,1). 4. For sum = 5: There are 4 pairs, (1,4), (2,3), (3,2), and (4,1). 5. For sum = 6: There are 5 pairs, (1,5), (2,4), (3,3), (4,2), and (5,1). 6. For sum = 7: There are 6 pairs, (1,6), (2,5), (3,4), (4,3), (5,2), and (6,1). 7. For sum = 8: There are 5 pairs, (2,6), (3,5), (4,4), (5,3), and (6,2). 8. For sum = 9: There are 4 pairs, (3,6), (4,5), (5,4), and (6,3). 9. For sum = 10: There are 3 pairs, (4,6), (5,5), and (6,4). 10. For sum = 11: There are 2 pairs, (5,6) and (6,5). 11. For sum = 12: There is 1 pair, (6,6).

Calculate the probability for each sum from 2 to 12

Now, we will calculate the probability of obtaining each sum by dividing the number of favorable outcomes for that sum by the total number of possible outcomes: 1. Probability of sum = 2: \( \frac{1}{36} \) 2. Probability of sum = 3: \( \frac{2}{36} = \frac{1}{18} \) 3. Probability of sum = 4: \( \frac{3}{36} = \frac{1}{12} \) 4. Probability of sum = 5: \( \frac{4}{36} = \frac{1}{9} \) 5. Probability of sum = 6: \( \frac{5}{36} \) 6. Probability of sum = 7: \( \frac{6}{36} = \frac{1}{6} \) 7. Probability of sum = 8: \( \frac{5}{36} \) 8. Probability of sum = 9: \( \frac{4}{36} = \frac{1}{9} \) 9. Probability of sum = 10: \( \frac{3}{36} = \frac{1}{12} \) 10. Probability of sum = 11: \( \frac{2}{36} = \frac{1}{18} \) 11. Probability of sum = 12: \( \frac{1}{36} \) The final probabilities are as follows: P(sum=2) = \(\frac{1}{36}\), P(sum=3) = \(\frac{1}{18}\), P(sum=4) = \(\frac{1}{12}\), P(sum=5) = \(\frac{1}{9}\), P(sum=6) = \(\frac{5}{36}\), P(sum=7) = \(\frac{1}{6}\), P(sum=8) = \(\frac{5}{36}\), P(sum=9) = \(\frac{1}{9}\), P(sum=10) = \(\frac{1}{12}\), P(sum=11) = \(\frac{1}{18}\), P(sum=12) = \(\frac{1}{36}\).

Key concepts

Combinatorial Probability

Combinatorial probability involves analyzing scenarios where outcomes are discrete and can be counted. It's a branch of mathematics that deals with counting and arrangement possibilities, and it is essential for determining the likelihood of certain events occurring.

When we throw two dice, each die has six faces, and each face is equally likely to come up, so there are a total of 6 x 6 = 36 possible outcomes. These outcomes are discrete; you can list them out one by one. Now, to determine the probability of a specific sum arising, we count the number of favorable outcomes that correspond to that sum (as was done in our step-by-step solution) and divide by the total number of possible combinations.

The process of counting these favorable outcomes for a specific event is a fundamental aspect of combinatorial probability. In the context of dice, when we want to know the chances of getting a sum of, say, 7, we're engaging in a classic combinatorial probability exercise. By improving our understanding of how to list combinations and permutations, we can more accurately calculate probabilities of these and similar events.

Probabilistic Models

Probabilistic models are mathematical representations that allow us to quantify the likelihood of different outcomes in a random process. In the context of dice games, our model assumes that the dice are fair, which means each of the six numbers is equally likely to occur on a single roll.

In our exercise, the model starts with listing all the 36 possible outcomes. This exhaustive enumeration is a precursor to identifying the elements that make up the event we're interested in. In such a model, probabilities are assigned to outcomes based on the assumption of fairness: since all outcomes are equally likely, each combination has a probability of 1/36.

To calculate the probability of a specific sum, we then look at how many outcomes result in that sum. Our model suggests the probability for the sum 'i' is the number of times 'i' appears in our list divided by the total number of outcomes. This can then be scaled to various other scenarios in the real world, beyond the roll of dice, where outcomes are not equiprobable, and a more complex probabilistic model would be required.

Dice Probability Distribution

Dice probability distribution describes how the probabilities are spread out over all possible sums of two dice. Different sums can have varying probabilities of occurring based on combinatorial principles. This distribution is a concrete example of a discrete probability distribution, which shows the probability for each possible discrete outcome.

In our exercise, the distribution is not uniform. For instance, there is only one way to get a sum of 2 (rolling a 1 on both dice), making it less probable than a sum of 7, where there are six combinations that result in this sum. This is why P(sum=2) is 1/36 while P(sum=7) is 1/6, making the distribution skewed towards the center sums.

The distribution peaks at 7, where it's most probable to roll this sum and tapers off towards the extremes (2 and 12). Understanding this distribution helps with grasping that not all outcomes in a random event have the same chance of occurring, and some can be significantly more likely than others. This understanding is crucial when predicting probabilities in games, decision-making, or any situation involving random outcomes.