Math Problem Statement

Bei jedem Laplace-Spielwürfel tritt die Augenzahl 6 mit der Wahrscheinlichkeit 1/6 auf. Thomas glaubt nur dann, dass ein vorliegender Spielwürfel ein Laplace-Spielwürfel ist, wenn bei 100 Würfen mit diesem Würfel die Anzahl der Sechsen mindestens 13 und höchstens 21 beträgt. Wie groß ist bei diesem Test die Wahrscheinlichkeit, dass Thomas einen Laplace-Spielwürfel nicht als solchen erkennt?

Solution

This problem revolves around a statistical test to assess whether a die follows the Laplace model (fair dice), based on the frequency of rolling a six over 100 trials. The main steps involve:

  1. Modeling the probabilities: Using the binomial distribution B(n=100,p=1/6)B(n=100, p=1/6), as the probability of rolling a six on one throw is p=1/6p = 1/6.
  2. Condition check: Thomas concludes the die is fair only if the number of sixes lies between 13 and 21, inclusive.
  3. Calculation of the required probability: Determine P(13X21)P(13 \leq X \leq 21), where XX represents the number of sixes rolled.

Let me solve it step by step:


Step 1: Probability Distribution

The random variable XX, the number of sixes rolled in 100 trials, follows a binomial distribution: XB(n=100,p=1/6)X \sim B(n=100, p=1/6) The probability mass function is: P(X=k)=(nk)pk(1p)nkP(X = k) = \binom{n}{k} p^k (1-p)^{n-k}


Step 2: Compute P(13X21)P(13 \leq X \leq 21)

This involves summing probabilities: P(13X21)=k=1321P(X=k)P(13 \leq X \leq 21) = \sum_{k=13}^{21} P(X = k)

For efficiency, we often use the normal approximation to the binomial distribution: XN(μ,σ2)X \approx N(\mu, \sigma^2) where: μ=np=1001616.67,σ=np(1p)=10016563.726\mu = np = 100 \cdot \frac{1}{6} \approx 16.67, \quad \sigma = \sqrt{np(1-p)} = \sqrt{100 \cdot \frac{1}{6} \cdot \frac{5}{6}} \approx 3.726

The range 13X2113 \leq X \leq 21 is approximated by: P(13X21)P(12.5μσZ21.5μσ)P(13 \leq X \leq 21) \approx P\left(\frac{12.5 - \mu}{\sigma} \leq Z \leq \frac{21.5 - \mu}{\sigma}\right)


Step 3: Standardization

Using the values of μ\mu and σ\sigma: P(13X21)P(12.516.673.726Z21.516.673.726)P(13 \leq X \leq 21) \approx P\left(\frac{12.5 - 16.67}{3.726} \leq Z \leq \frac{21.5 - 16.67}{3.726}\right) P(13X21)P(1.12Z1.30)P(13 \leq X \leq 21) \approx P(-1.12 \leq Z \leq 1.30)

From the standard normal table: P(Z1.30)0.9032,P(Z1.12)0.1314P(Z \leq 1.30) \approx 0.9032, \quad P(Z \leq -1.12) \approx 0.1314 Thus: P(13X21)0.90320.1314=0.7718P(13 \leq X \leq 21) \approx 0.9032 - 0.1314 = 0.7718


Final Answer

The probability that Thomas concludes the die is fair is approximately 77.18%.


Would you like to see the detailed binomial calculations or confirm this with a specific computation? Let me know! 😊


Related Questions

  1. What are the characteristics of the Laplace distribution for fair dice rolls?
  2. How does the binomial distribution apply to dice rolling experiments?
  3. Why is the normal approximation valid in this case?
  4. How would the conclusion change if the range 13X2113 \leq X \leq 21 were altered?
  5. What is the probability of rolling fewer than 13 sixes in this experiment?

Tip:

For binomial distributions with large nn and pp not too close to 0 or 1, the normal approximation simplifies calculations greatly!

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Math Problem Analysis

Mathematical Concepts

Probability Theory
Binomial Distribution
Normal Approximation

Formulas

P(X = k) = (n choose k) * p^k * (1-p)^(n-k)
μ = np
σ = √(np(1-p))
P(a ≤ X ≤ b) ≈ P((a-0.5 - μ)/σ ≤ Z ≤ (b+0.5 - μ)/σ)

Theorems

Central Limit Theorem
Properties of the Normal Distribution

Suitable Grade Level

Grade 11-12 (Advanced High School) or University Level