Book edition 1st

Author(s) Daniel V. Schroeder

Pages 356 pages

ISBN 9780201380279

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Short Answer

Suppose you flip $1000$ coins.
$(a)$ 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.) $(b)$ What is the probability of getting exactly $600$ heads and $400$ tails?

Part $(a)$

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

part $(b)$

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

See the step by step solution

Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

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

$\begin{array}{l} Ω (N, n) = (\begin{matrix} N \\ n \end{matrix}) = \frac{N!}{n! (N - n)!} \\ Ω (1000, 500) = (\begin{matrix} 1000 \\ 500 \end{matrix}) \\ Ω (1000, 500) = \frac{1000!}{500! (1000 - 500)!} \\ Ω (1000, 500) = \frac{1000!}{500!^{2}} . \end{array}$

Using Stirling's approximation as

$\begin{array}{l} N! \approx N^{N} e^{- N} \sqrt{2 π N} \\ Ω (1000, 500) = \frac{1000!}{500!^{2}} \approx \frac{1000^{1000} e^{- 1000} \sqrt{2 π \cdot 1000}}{{(500^{500} e^{- 500} \sqrt{2 π \cdot 500})}^{2}} \\ Ω (1000, 500) \approx \frac{1000^{1000} \times e^{- 1000} \times \sqrt{1000}}{500^{1000} \times e^{- 1000} \times 500 \times \sqrt{2 π}} \end{array}$

Step:2 Dervative: (part a)

From the above equation,

$\begin{array}{l} Ω (1000, 500) \approx \frac{(500 \times 2)^{1000} \times \sqrt{1000}}{500^{1000} \times 500 \times \sqrt{2 π}} \\ Ω (1000, 500) \approx \frac{(2)^{1000} \times \sqrt{1000}}{500 \times \sqrt{2 π}} \end{array}$

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

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

Substituting,we get

$\begin{array}{l} P (500) = \frac{Ω (1000, 500)}{2^{1000}} \approx \frac{1}{2^{1000}} \frac{(2)^{1000} \times \sqrt{1000}}{500 \times \sqrt{2 π}} \\ P (500) \approx \frac{\sqrt{1000}}{500 \times \sqrt{2 π}} \\ P (500) \approx 0.02523 . \end{array}$

Step: 3 Derivative probability: (part b)

The number getting $n = 500$ heads as

$\begin{array}{l} Ω (N, n) = (\begin{matrix} N \\ n \end{matrix}) \\ Ω (N, n) = \frac{N!}{n! (N - n)!} \\ Ω (1000, 600) = (\begin{matrix} 1000 \\ 600 \end{matrix}) \\ Ω (1000, 600) = \frac{1000!}{600! (1000 - 600)!} \\ Ω (1000, 600) = \frac{1000!}{600! 400!} . \end{array}$

Using Stirling's approximation as

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

Step: 4 Equating part: (part b)

From the above equation,

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

Probability approximately getting exactly $500$ heads as

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

Substituting,we get

$\begin{array}{l} P (600) = \frac{Ω (1000, 600)}{2^{1000} \approx \frac{1}{2^{1000}}} \times \sqrt{\frac{1}{480 π}} \times \frac{2^{1000} \times 500^{1000}}{400^{400} \times 600^{600}} \\ P (600) \approx \sqrt{\frac{1}{480 π}} \times \frac{500^{1000}}{400^{400} \times 600^{600}} \end{array}$

Step: 5 Finding probability value: (part b)

The ratio is not applicalable, so simplify as

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

But,

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

Where,

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

Where as Maple is working large exponents directly.

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