Math Problem Statement

Ten unbiased coins are tossed independently. Let X represent the total number of heads obtained from these tosses. Using Chebyshev's inequality, find an upper bound on the probability that the total number of heads deviates from the expected value of 5 by 2 or more. Enter the answer correct to two decimal places.

Solution

This problem requires the use of Chebyshev's inequality to find an upper bound on the probability that the number of heads deviates from the expected value by 2 or more.

Let’s break this down:

  1. The number of heads XX follows a binomial distribution: XBinomial(n=10,p=0.5)X \sim \text{Binomial}(n = 10, p = 0.5) where n=10n = 10 is the number of trials (coin flips), and p=0.5p = 0.5 is the probability of getting a head in each flip.

  2. The expected value μ\mu of XX is: μ=np=10×0.5=5\mu = np = 10 \times 0.5 = 5

  3. The variance σ2\sigma^2 of XX is: σ2=np(1p)=10×0.5×0.5=2.5\sigma^2 = np(1 - p) = 10 \times 0.5 \times 0.5 = 2.5 So, the standard deviation σ\sigma is: σ=2.51.5811\sigma = \sqrt{2.5} \approx 1.5811

  4. We need to find the probability that XX deviates from its mean by 2 or more: P(X52)P(|X - 5| \geq 2)

    Using Chebyshev’s inequality: P(Xμkσ)1k2P(|X - \mu| \geq k\sigma) \leq \frac{1}{k^2} Here, the deviation X52|X - 5| \geq 2 corresponds to k=2σk = \frac{2}{\sigma}.

    Substituting σ=1.5811\sigma = 1.5811: k=21.58111.265k = \frac{2}{1.5811} \approx 1.265

  5. Finally, applying Chebyshev’s inequality: P(X52)1(1.265)211.5990.6255P(|X - 5| \geq 2) \leq \frac{1}{(1.265)^2} \approx \frac{1}{1.599} \approx 0.6255

Therefore, the upper bound on the probability is approximately 0.63.


Would you like more details on any part of this solution? Here are five related questions to deepen your understanding:

  1. How would the upper bound change if the deviation was 3 instead of 2?
  2. What does Chebyshev's inequality tell us about distributions beyond the binomial?
  3. How could you calculate the exact probability using the binomial distribution?
  4. Can Chebyshev's inequality be applied to any distribution? Why or why not?
  5. What is the significance of the standard deviation in Chebyshev’s inequality?

Tip: Chebyshev’s inequality gives a conservative estimate, meaning the actual probability might be lower than the bound. It's useful when little is known about the distribution.

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Math Problem Analysis

Mathematical Concepts

Probability
Binomial Distribution
Chebyshev's Inequality
Variance

Formulas

Expected value: E(X) = np
Variance: Var(X) = np(1 - p)
Chebyshev's inequality: P(|X - μ| ≥ kσ) ≤ 1/k²

Theorems

Chebyshev's Inequality

Suitable Grade Level

Grades 10-12