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Suppose you flip 50fair coins.

(a) How many possible outcomes (microstates) are there?

(b) How many ways are there of getting exactly25heads and25tails?

(c) What is the probability of getting exactly 25heads and 25tails?

(d) What is the probability of getting exactly 30heads and 20tails?

(e) What is the probability of getting exactly 40heads and 10 tails?

(f) What is the probability of getting 50heads and no tails?

(g) Plot a graph of the probability of getting n heads, as a function of n.

Short Answer

Expert verified

(a)Possible outcomes (microstates) are there 1.26411014microstate,

(b) The ways are there of getting exactly 25heads and25tails are 1.26411014microstate

(c)The is the probability of getting exactly 25heads and 25tails are P=0.1123

(d) The probability of getting exactly 30heads and 20tails? P=0.04186

(e) The probability of getting exactly 40heads and 10tails? P=9.1236106

(f)The probability of getting 50heads and no tails are P=8.8821016

(e) A graph of the probability of getting n heads, as a function of n.

Step by step solution

01

Step1:Explanation of 50 fair coins (part a b c and d)

(a) we suppose we flip 50coins, n=50. The total number of microstates is:

2n=250=1.12591015microstate

(b) The number of microstates receiving 25heads is as follows:

Nn

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

5025=50!25!(5025)!=1.26411014microstate

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

P=NnTotal number of microstates

P=5025250=0.1123

(d) The probability of getting the macrostate of30heads and 20tails, regardless of order, is:

P=5030250=50!30!(5030)!250=0.04186

02

Step2:Explanation of  50 fair coins(part e f and  g )

(e)The probability of obtaining the macrostate of40heads and10tails, regardless of order, is as follows:

localid="1650287550141" P=5040250=50!40!(5040)!250=9.1236106

(f) Regardless of order, the probability of receiving the macrostate of50heads and 0tails is:

localid="1650287553981" P=500250=50!0!(500)!250=8.8821016

(g)The function we need to plot is the number of heads divided by the probability, so is:

localid="1650287557923" P(n)=(50,n)250=50!n!(50n)!250

The table shows the values that we intend to plot where n varies from0to50. I drew the graph in MS Excel, and the image below shows how I generated the data.

03

Step3:Probability of 50 fair coins

04

Step4:find graph

The probability decreases sharply as we move away from the midpoint, with an equal number of heads and tails. This is nicely illustrated by a plot of the probability P(N):

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Most popular questions from this chapter

Use a computer to plot formula 2.22directly, as follows. Define z=q_{A}/q, so that (1-z)=q_{B}/q. Then, aside from an overall constant that we'll ignore, the multiplicity function is [4z(1-z)]N, where zranges from 0to1and the factor of 4ensures that the height of the peak is equal to 1for any N. Plot this function forN=1,10,100,1000, and 10,000. Observe how the width of the peak decreases asNincreases.

The natural logarithm function, ln, is defined so that elnx=xfor any positive numberx.
aSketch a graph of the natural logarithm function.
b Prove the identities
localid="1650331641178" lnab=lna+lnbandlocalid="1650331643409" lnab=blna
(c) Prove thatlocalid="1650331645612" ddxlnx=1x.
(d) Derive the useful approximation

localid="1650331649052" ln(1+x)x

which is valid when localid="1650331651790" |x|1. Use a calculator to check the accuracy of this approximation for localid="1650331654235" x=0.1and localid="1650331656447" x=0.01.

For a single large two-state paramagnet, the multiplicity function is very sharply peaked about N=N/2.

(a) Use Stirling's approximation to estimate the height of the peak in the multiplicity function.

(b) Use the methods of this section to derive a formula for the multiplicity function in the vicinity of the peak, in terms of xN(N/2). Check that your formula agrees with your answer to part (a) when x=0.

(c) How wide is the peak in the multiplicity function?

(d) Suppose you flip 1,000,000coins. Would you be surprised to obtain heads and 499,000 tails? Would you be surprised to obtain 510,000 heads and 490,000 tails? Explain.

Use Stirling's approximation to show that the multiplicity of an Einstein solid, for any large values ofNandlocalid="1650383388983" q,is approximately

Omega(N,q)q+Nqqq+NNN2q(q+N)/N

The square root in the denominator is merely large, and can often be neglected. However, it is needed in Problem2.22. (Hint: First show that=Nq+N(q+N)!q!N!. Do not neglect the2Nin Stirling's approximation.)

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