(a) Suppose that Martian dice are regular tetrahedra with vertices labeled 1 to 4 Two such dice are tossed and the sum of the numbers showing is even. Let \(x\) be this sum. Set up the sample space for \(x\) and the associated probabilities. (b) Find \(E(x)\) and \(\sigma_{x}\) (c) Find the probability of exactly fifteen 2 's in 48 tosses of a Martian die using the binomial distribution. (d) Approximate (c) using the normal distribution. (e) Approximate (c) using the Poisson distribution.

The sample space is a fundamental concept in probability theory. It represents all possible outcomes of an experiment. In the context of our exercise with Martian dice, the sample space is crucial for understanding the various sums we can get from rolling the dice. Each Martian die has numbers from 1 to 4. When two dice are tossed, the possible sums of the numbers showing range from 2 to 8. If we consider only the even sums, the possible values are 2, 4, 6, and 8. In our solution, we carefully counted the number of ways each even sum can occur:

- Sum 2: (1,1) - 1 outcome
- Sum 4: (1,3), (2,2), (3,1) - 3 outcomes
- Sum 6: (2,4), (3,3), (4,2) - 3 outcomes
- Sum 8: (4,4) - 1 outcome
Understanding the sample space allows us to calculate the probabilities for each of these sums.

Expected value is an important concept in probability and statistics. It represents the average or mean value of a random variable over many trials of an experiment. Calculating the expected value involves multiplying each possible value of the random variable by its probability and summing these products. For our dice example, the expected value of the sum of the dice rolls can be found using:e(x) = ∑ [x * P(x)]where x represents the sum, and P(x) is the probability of x. Based on our earlier calculations for sums 2, 4, 6, and 8, the expected value is computed as:e(x) = 2*(1/16) + 4*(3/16) + 6*(3/16) + 8*(1/16) = 2.5This tells us that, on average, the sum of two Martian dice when the sum is even will be 2.5.

The binomial distribution describes the number of successes in a fixed number of independent trials of an experiment, where each trial has the same probability of success. In our problem, we consider the probability of getting exactly fifteen 2's in 48 tosses of a Martian die. Each toss is independent and has a success probability (getting a 2) of 1/4. The binomial probability formula is: P(X = k) = C(n, k) * p^k * (1-p)^(n-k)where n is the number of trials, k is the number of successes, and p is the probability of success. For our example, we set n = 48, k = 15, and p = 0.25 and calculate:P(X = 15) = C(48, 15) * (0.25)^15 * (0.75)^(33)This can be a bit complicated to compute by hand but is straightforward using statistical software or a calculator.

The normal distribution, often called the Gaussian distribution, is a continuous probability distribution. It is useful for approximating the binomial distribution when the number of trials n is large and the probability of success p is neither very small nor very large. For our problem, we can use the normal distribution to approximate the probability of exactly fifteen 2's in 48 tosses of a Martian die. We start by calculating the mean (μ) and variance (σ²) of our binomial distribution: μ = np = 48 * 0.25 = 12σ² = np(1-p) = 48 * 0.25 * 0.75 = 9σ = √9 = 3Using the continuity correction factor, we approximate P(X=15) by converting it to a z-score: z = (X - μ) / σ = (14.5 - 12) / 3 ≈ 0.833We would then use the standard normal distribution table to find the cumulative probability for this z-score.

The Poisson distribution is used for modeling the number of events occurring within a fixed interval of time or space when these events happen with a known constant mean rate and independently of the time since the last event. In our exercise, we approximate the probability of getting exactly fifteen 2's in 48 tosses of a Martian die using the Poisson distribution. The Poisson distribution is characterized by the parameter λ (lambda), which represents the mean number of occurrences within the given interval. In our example: λ = np = 48 * 0.25 = 12The probability of observing k events in an interval is given by: P(X = k) = (λ^k * e^(-λ)) / k!For k = 15, λ = 12, approximating by: P(X=15) ≈ (12^15 * e^(-12)) / 15!Like the binomial probability, this might require computational tools for precise calculation.