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Q. 2.16

Expert-verifiedFound in: Page 63

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 $P\left(600\right)\approx 4.635\times {10}^{-11}.$

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

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

Using Stirling's approximation as

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

From the above equation,

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

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

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

Substituting,we get

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

The number getting $n=500$ heads as

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

Using Stirling's approximation as

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

From the above equation,

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

Probability approximately getting exactly $500$ heads as

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

Substituting,we get

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

The ratio is not applicalable, so simplify as

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

But,

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

Where,

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

Where as Maple is working large exponents directly.

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