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Expert-verified Found in: Page 51 ### An Introduction to Thermal Physics

Book edition 1st
Author(s) Daniel V. Schroeder
Pages 356 pages
ISBN 9780201380279 # Suppose you flip $20$ fair coins.(a) How many possible outcomes (microstates) are there?(b) What is the probability of getting the sequence HTHHTTTHTHHHTHHHHTHT (in exactly that order)?(c) What is the probability of getting $12$ heads and $8$ tails (in any order)?

(a)The possible outcomes microstate are$\text{}1048576\text{microstates}$

(b) The probability of getting the sequence is$P=9.536×{10}^{-7}$

(c)The probability of getting 12 heads and 8 tails are$\text{}P=0.12$

See the step by step solution

## Step1:possible outcome of macrostate(part a)

(a)We suppose we flip 20 coins, $n=20$. The total number of microstates will be:

Calculation method

${2}^{n}={2}^{20}$

$=1048576$microstate

So, the answer is ${2}^{20}$ or 1048576 microstates.

## Step2:The probability of sequence(part b)

(b)The likelihood of obtaining any specific sequence of heads and tails (for example HTHHTTTHTHHHTHHHHTHT) is

$P=\frac{\text{Favorable microstate}}{\text{Total number of microstates}}$

$\begin{array}{r}=\frac{1}{{2}^{n}}\\ =\frac{1}{{2}^{20}}\end{array}$

$=\frac{1}{1048576}$

$=9.536×{10}^{-7}$

As a result, the answer is for the sequence of heads and tails. (for example HTHHTTTHTHHHTHHHHTHT) is $\frac{1}{{2}^{20}}$ or $9.536×{10}^{-7}$.

## Step3:Probability of 12 heads and 8 tails(part c)

(c)Regardless of order, the probability of getting the macrostate of 12 heads and 8 tails is given by:

$P=\frac{\left(\begin{array}{l}N\\ n\end{array}\right)}{\text{Total number of microstates}}$

where n denotes the number of heads and N denotes the number of coins, so:

$P=\frac{\left(\begin{array}{l}20\\ 12\end{array}\right)}{1048576}$

After arranging we get,

$\left(\begin{array}{l}20\\ 12\end{array}\right)=\frac{20!}{12!\left(20-12\right)!}=125970$

Finally solution is

$P=\frac{125970}{1048576}=0.12$ ### Want to see more solutions like these? 