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Chapter 1: Problem 7

A fairly-weighted coin is tossed 20 times. a. Prove that the probability for a specific number of heads is given by the binomial distribution. b. Calculate the probabilities for each of the possible numbers of heads. c. Recalculate the probabilities in part (b) by invoking a Gaussian distribution. d. Display your results from parts (b) and (c) on a single plot. Discuss the implications of your comparison.

Short Answer

Expert verified
The binomial distribution can be used to represent discrete probabilities for a specific number of successes in trials and is given by the formula \( \binom{n}{k}p^k(1-p)^{n-k} \). For a fair coin flipped 20 times, you calculate probabilities for k = 0 to 20 using the binomial distribution formula, then you use the Gaussian approximation and compare both on a plot. The Gaussian is an approximation and might not fit perfectly for smaller values of n.

Step by step solution

01

Understanding the Binomial Distribution

A binomial distribution is appropriate for a sequence of independent experiments, each with two outcomes (such as heads or tails in a coin toss) with the probability of success (getting heads) the same on each trial. This distribution gives the probability of having exactly k successes (heads) in n trials (tosses). The formula for the binomial distribution is \[ P(k; n, p) = \binom{n}{k}p^k(1-p)^{n-k} \] where \( P(k; n, p) \) is the probability of k successes in n trials, \( \binom{n}{k} \) is the binomial coefficient, p is the probability of success, and (1-p) is the probability of failure (tails in this case).
02

Calculating Probabilities for Each Number of Heads

To calculate the probability of getting a specific number of heads (k) out of 20 tosses (n=20), use the given binomial formula. Since the coin is fairly weighted, the probability of getting heads (p) is 0.5. The probability for getting k heads is thus given by: \[ P(k; 20, 0.5) = \binom{20}{k}(0.5)^k(0.5)^{20-k} \] You would calculate this for each k from 0 to 20 to find all the probabilities.
03

Recalculating Using the Gaussian Distribution

Under certain conditions (large number of trials n), the binomial distribution can be approximated by the Gaussian (or normal) distribution. The Gaussian distribution is given by the probability density function: \[ P(x) = \frac{1}{\sqrt{2\pi}\sigma}e^{-\frac{{(x-\mu)^2}}{{2\sigma^2}}} \] where \( \mu = np \) is the mean and \( \sigma^2 = np(1-p) \) is the variance. For a fair coin, \( \mu = 20(0.5) = 10 \) and \( \sigma^2 = 20(0.5)(0.5) = 5 \). Calculate this for each integer value of x from 0 to 20 to find the approximate probabilities.
04

Plotting and Comparing the Results

To compare the calculations from the binomial and Gaussian distributions, use a plotting tool (like Matplotlib if doing this in Python) to create a bar chart for binomial probabilities and a line plot for Gaussian probabilities. Plot the number of heads (k or x) on the x-axis and the corresponding probabilities from both distributions on the y-axis.
05

Discussion of Implications

Evaluate the closeness of the two distributions by observing the plot curves. Normally, you would expect the Gaussian plot to be a good fit for the binomial probabilities when n is large. Discuss the accuracy of the Gaussian approximation for the binomial results in this specific case of n = 20.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Probability Theory
Probability theory is the branch of mathematics concerned with the analysis of random phenomena. The fundamental component of probability theory is the probability - a measure that quantifies the likelihood that a specific event will occur within a given set of outcomes.

When flipping a coin, as in our exercise, the occurrence of heads or tails is inherently random, yet probability theory allows us to calculate the expected occurrence of these events over time. The most basic probability model for a coin toss is a binomial distribution, where each toss is independent of the other and has only two possible outcomes. In our exercise, we determine the probability of obtaining a certain number of heads in 20 tosses — a classic example of a binomial scenario.

Understanding probability helps students reason quantitatively about uncertain scenarios, making it an invaluable tool in fields ranging from finance to science, and even everyday decision making.
Gaussian Distribution
The Gaussian distribution, also known as the normal distribution, is a continuous probability distribution that is symmetric around its mean, representing a common occurrence of values near this mean. This distribution is characterized by its bell-shaped curve and is defined by the probability density function:
\[ P(x) = \frac{1}{\sqrt{2\pi}\sigma}e^{-\frac{{(x-\mu)^2}}{{2\sigma^2}}} \]
where \( \mu \) is the mean and \( \sigma^2 \) is the variance. What makes the Gaussian distribution extremely useful and ubiquitous in statistics is its ability to describe many natural phenomena, due to the Central Limit Theorem, which states that the sum (or average) of independent random variables tends toward a Gaussian distribution, regardless of the original distribution of the variables.

In our exercise, the Gaussian distribution is used as an approximation for the binomial distribution when the number of trials is large, which simplifies calculations and helps in understanding the underlying patterns of random events such as the flip of a coin.
Probability Distribution Comparison
Comparing probability distributions, like the binomial and Gaussian distributions, is a foundational concept in statistics that allows for the understanding of the relationships between different types of distributions and how well they approximate one another in various conditions.

In the step-by-step exercise, we encounter such a comparison. After calculating the individual probabilities of head occurrences using the binomial distribution, we then approximate these probabilities using the Gaussian distribution. This comparison is visually represented on a plot, showcasing the fit of the Gaussian to the binomial, especially as the number of trials grows. Large sample sizes generally lead to the binomial probabilities converging on the Gaussian curve due to the aforementioned Central Limit Theorem.

Practical implications of this comparison are vast, from making predictions in sports analysis to assessing risks in financial portfolios. Understanding how well different distributions approximate each other informs decision-making and helps in developing a more intuitive sense of probability distributions.

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Most popular questions from this chapter

Gaussian integrals appear often in statistical thermodynamics and particularly in kinetic theory. Consider the zeroth order Gaussian integral \(I_{0}(\alpha)\) and the gamma function \(\Gamma(x)\) : $$ I_{0}(\alpha)=\int_{0}^{\infty} e^{-\alpha x^{2}} d x \quad \Gamma(x)=\int_{0}^{\infty} t^{x-1} e^{-t} d t . $$ a. Evaluate \(I_{0}(\alpha)\) by squaring it, i.e., $$ I_{0}^{2}=\int_{0}^{\infty} \int_{0}^{\infty} e^{-\alpha x^{2}} e^{-\alpha y^{2}} d x d y, $$ and transforming to polar coordinates. b. Demonstrate the following results for the gamma function: (1) \(\Gamma(1)=1\) (2) \(\Gamma(x+1)=x \Gamma(x)\) (3) \(\Gamma(n+1)=n\) ! for \(n\) an integer. c. Show that \(\Gamma(1 / 2)=\sqrt{\pi}\). d. Verify the following standard expression for the \(n\)th order Gaussian integral: $$ I_{n}(\alpha)=\int_{0}^{\infty} x^{n} e^{-\alpha x^{2}} d x=\frac{1}{2 \alpha^{(n+1) / 2}} \Gamma\left(\frac{n+1}{2}\right) . $$ Use this expression to evaluate the Gaussian integrals for \(n=0-5\). Compare your results to those tabulated in Appendix \(\mathrm{B}\).

Let \(f(x, y)\) be a joint probability density function, i.e., \(f(x, y) d x d y\) is the probability that \(X\) lies between \(x\) and \(x+d x\) and \(Y\) lies between \(y\) and \(y+d y\). If \(X\) and \(Y\) are independent, then $$ f(x, y) d x d y=f_{1}(x) f_{2}(y) d x d y . $$ Therefore, if \(W=X+Y\), show that $$ \bar{W}=\bar{X}+\bar{Y} $$ and that $$ \overline{(W-\bar{W})^{2}}=\overline{(X-\bar{X})^{2}}+\overline{(Y-\bar{Y})^{2}} . $$ In other words, if \(X\) and \(Y\) are independent, the mean and variance of their sum is equal to the sum of their means and variances.

Consider the probability function for the binomial distribution, $$ B(M)=\frac{N !}{M !(N-M) !} p^{M} q^{N-M}, $$ where \(N\) and \(p\) are specified, and the probability \(q=1-p\). a. Show that the distribution is properly normalized. b. Verify that the mean \(\bar{M}=N p\). c. Show that the standard deviation \(\sigma=\sqrt{N p(1-p)}\). d. The fluctuation is defined as the ratio of the standard deviation to the mean. Determine the fluctuation of the binomial distribution for \(p=0.5\). What happens to the fluctuation for large values of \(N\) ? What are the physical implications of this result when considering the thermodynamic properties of an ideal gas? Hint: Recall that, for arbitrary values of \(p\) and \(q\), the binomial theorem gives $$ \sum_{M=0}^{N} B(M)=(p+q)^{N} $$ so that, for any value of \(s\), $$ \sum_{M=0}^{N} s^{M} B(M)=(s p+q)^{N} . $$ The trick now is to differentiate with respect to \(s\).

Assuming that the inherent ratio of male to female children is unity, determine the following probabilities for a family of six children. a. The four oldest children will be boys and the two youngest will be girls. b. Exactly half the children will be boys. c. All six children will be of the same sex. d. A second girl is born last.
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