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Answers without the blur. Sign up and see all textbooks for free! Q. 2.16

Expert-verified Found in: Page 63 ### An Introduction to Thermal Physics

Book edition 1st
Author(s) Daniel V. Schroeder
Pages 356 pages
ISBN 9780201380279 # Suppose you flip $1000$ coins. $\left(a\right)$ What is the probability of getting exactly $500$heads and $500$ tails? (Hint: First write down a formula for the total number of possible outcomes. Then, to determine the "multiplicity" of the $500-500$ "macrostate," use Stirling's approximation. If you have a fancy calculator that makes Stirling's approximation unnecessary, multiply all the numbers in this problem by $10$ , or $100$ , or $1000$, until Stirling's approximation becomes necessary.) $\left(b\right)$ What is the probability of getting exactly $600$ heads and $400$ tails?

Part $\left(a\right)$

$\left(a\right)$The probability of getting exactly $P\left(500\right)\approx 0.02523.$

part $\left(b\right)$

$\left(b\right)$he probability of getting exactly $P\left(600\right)\approx 4.635×{10}^{-11}.$

See the step by step solution

## Step: 1 Finding probability: (part a)

A larger coin tossing experiment is underway. N=1000 coins are flipped. Stirling's approach may be used to calculate the likelihood of receiving exactly 500 head and 500 tails. The total number of possible outcomes is as follows:

${2}^{N}={2}^{1000}$

Where $n=500$ heads as

$\begin{array}{l}\mathrm{\Omega }\left(N,n\right)=\left(\begin{array}{c}N\\ n\end{array}\right)=\frac{N!}{n!\left(N-n\right)!}\\ \mathrm{\Omega }\left(1000,500\right)=\left(\begin{array}{c}1000\\ 500\end{array}\right)\\ \mathrm{\Omega }\left(1000,500\right)=\frac{1000!}{500!\left(1000-500\right)!}\\ \mathrm{\Omega }\left(1000,500\right)=\frac{1000!}{500{!}^{2}}.\end{array}$

Using Stirling's approximation as

$\begin{array}{l}N!\approx {N}^{N}{e}^{-N}\sqrt{2\pi N}\\ \mathrm{\Omega }\left(1000,500\right)=\frac{1000!}{500{!}^{2}}\approx \frac{{1000}^{1000}{e}^{-1000}\sqrt{2\pi \cdot 1000}}{{\left({500}^{500}{e}^{-500}\sqrt{2\pi \cdot 500}\right)}^{2}}\\ \mathrm{\Omega }\left(1000,500\right)\approx \frac{{1000}^{1000}×{e}^{-1000}×\sqrt{1000}}{{500}^{1000}×{e}^{-1000}×500×\sqrt{2\pi }}\end{array}$

## Step:2 Dervative: (part a)

From the above equation,

$\begin{array}{l}\mathrm{\Omega }\left(1000,500\right)\approx \frac{\left(500×2{\right)}^{1000}×\sqrt{1000}}{{500}^{1000}×500×\sqrt{2\pi }}\\ \mathrm{\Omega }\left(1000,500\right)\approx \frac{\left(2{\right)}^{1000}×\sqrt{1000}}{500×\sqrt{2\pi }}\end{array}$

Probability approximately getting exactly role="math" localid="1650333432268" $500$heads as

$\begin{array}{l}P\left(n\right)=\frac{\mathrm{\Omega }\left(N,n\right)}{{2}^{N}}\\ P\left(500\right)=\frac{\mathrm{\Omega }\left(1000,500\right)}{{2}^{1000}}\end{array}$

Substituting,we get

$\begin{array}{l}P\left(500\right)=\frac{\mathrm{\Omega }\left(1000,500\right)}{{2}^{1000}}\approx \frac{1}{{2}^{1000}}\frac{\left(2{\right)}^{1000}×\sqrt{1000}}{500×\sqrt{2\pi }}\\ P\left(500\right)\approx \frac{\sqrt{1000}}{500×\sqrt{2\pi }}\\ P\left(500\right)\approx 0.02523.\end{array}$

## Step: 3 Derivative probability: (part b)

The number getting $n=500$ heads as

$\begin{array}{l}\mathrm{\Omega }\left(N,n\right)=\left(\begin{array}{c}N\\ n\end{array}\right)\\ \mathrm{\Omega }\left(N,n\right)=\frac{N!}{n!\left(N-n\right)!}\\ \mathrm{\Omega }\left(1000,600\right)=\left(\begin{array}{c}1000\\ 600\end{array}\right)\\ \mathrm{\Omega }\left(1000,600\right)=\frac{1000!}{600!\left(1000-600\right)!}\\ \mathrm{\Omega }\left(1000,600\right)=\frac{1000!}{600!400!}.\end{array}$

Using Stirling's approximation as

role="math" localid="1650333713655" $\begin{array}{l}N!\approx {N}^{N}{e}^{-N}\sqrt{2\pi N}\\ \mathrm{\Omega }\left(1000,600\right)=\frac{1000!}{600!400!}\approx \frac{{1000}^{1000}{e}^{-1000}\sqrt{2\pi ×1000}}{\left({600}^{600}{e}^{-600}\sqrt{2\pi ×600}\right)\left({400}^{400}{e}^{-400}\sqrt{2\pi ×\cdot 400}\right)}\\ \mathrm{\Omega }\left(1000,600\right)\approx \frac{{1000}^{1000}×{e}^{-1000}×\sqrt{1000}}{{400}^{400}×{600}^{600}×{e}^{-1000}×\sqrt{600×400}×\sqrt{2\pi }}\end{array}$

## Step: 4 Equating part: (part b)

From the above equation,

$\begin{array}{l}\mathrm{\Omega }\left(1000,600\right)\approx \frac{{2}^{1000}×{500}^{1000}×\sqrt{1000}}{{400}^{400}×{600}^{600}×\sqrt{600×400}×\sqrt{2\pi }}\\ \mathrm{\Omega }\left(1000,600\right)\approx \sqrt{\frac{1}{480\pi }}×\frac{{2}^{1000}×{500}^{1000}}{{400}^{400}×{600}^{600}}\end{array}$

Probability approximately getting exactly $500$ heads as

$\begin{array}{l}P\left(n\right)=\frac{\mathrm{\Omega }\left(N,n\right)}{{2}^{N}}\\ P\left(600\right)=\frac{\mathrm{\Omega }\left(1000,600\right)}{{2}^{1000}}\end{array}$

Substituting,we get

$\begin{array}{l}P\left(600\right)=\frac{\mathrm{\Omega }\left(1000,600\right)}{{2}^{1000}\approx \frac{1}{{2}^{1000}}}×\sqrt{\frac{1}{480\pi }}×\frac{{2}^{1000}×{500}^{1000}}{{400}^{400}×{600}^{600}}\\ P\left(600\right)\approx \sqrt{\frac{1}{480\pi }}×\frac{{500}^{1000}}{{400}^{400}×{600}^{600}}\end{array}$

## Step: 5 Finding probability value: (part b)

The ratio is not applicalable, so simplify as

$\begin{array}{l}\sqrt{\frac{1}{480\pi }}×\frac{{500}^{1000}}{{400}^{400}×{600}^{600}}=\sqrt{\frac{1}{480\pi }}×\frac{\left(1.25×400{\right)}^{1000}}{\left(1×400{\right)}^{400}×\left(1.5×400{\right)}^{600}}\\ \to \sqrt{\frac{1}{480\pi }}×\frac{\left(400{\right)}^{1000}×\left(1.25{\right)}^{1000}}{\left(400{\right)}^{400}×\left(1{\right)}^{400}×\left(400{\right)}^{600}×\left(1.5{\right)}^{600}}\end{array}$

But,

$\begin{array}{l}\frac{\left(400{\right)}^{1000}}{\left(400{\right)}^{600}×\left(400{\right)}^{400}}=\frac{\left(400{\right)}^{1000}}{\left(400{\right)}^{1000}}=1\\ \to \sqrt{\frac{1}{480\pi }}×\frac{\left(1.25{\right)}^{1000}}{\left(1{\right)}^{400}×\left(1.5{\right)}^{600}}\end{array}$

Where,

$\begin{array}{l}P\left(600\right)\approx \sqrt{\frac{1}{480\pi }}×\frac{\left(1.25{\right)}^{1000}}{\left(1{\right)}^{400}×\left(1.5{\right)}^{600}}\\ P\left(600\right)\approx 4.635×{10}^{-11}.\end{array}$

Where as Maple is working large exponents directly.

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