91Ó°ÊÓ

The number of ways ${\mathbf{W}}_{{\mathbf{N}}_{\mathbf{R}}}^{\mathbf{N}}$that$\mathbf{N}$particles can be arranged such that there is an imbalance of${\mathbf{N}}_{\mathbf{R}}$particles is given by the binomial coefficient:${\mathbf{W}}_{{\mathbf{N}}_{\mathbf{R}}}^{\mathbf{N}}\mathbf{=}\left(\begin{array}{c}N\\ N_R\end{array}\right)\mathbf{=}\frac{\mathbf{N}\mathbf{!}}{{\mathbf{N}}_{\mathbf{R}}\mathbf{!}\mathbf{(}\mathbf{N}\mathbf{−}{\mathbf{N}}_{\mathbf{R}}\mathbf{)}\mathbf{!}}\mathbf{.}$

a)

Before calculating the probabilities, we first have to calculate the number of ways for particles to be arranged with an imbalance of${\mathrm{N}}_{\mathrm{R}}$ particles. Since we are given a two-sided room, the number of ways${\mathrm{W}}_{{\mathrm{N}}_{\mathrm{R}}}^{\mathrm{N}}$ that $N$particles can be arranged such that there is an imbalance of$N_R$ particles is given by the binomial coefficient:

$W_{N_R}^N=\left(\begin{array}{c}N\\ N_R\end{array}\right)=\frac{N!}{N_R!(N−N_R)!}.$

Let us first consider the case$N=20$ .

Plugging in $N=20$, and

$N_R=\frac{1}{2}N+5=\frac{1}{2}20+5=15$ into eq. (1) we obtain:

$\begin{array}{c}{\mathrm{W}}_{15}^{20}=\frac{20!}{15!(20−15)!}\\ =1.5×{10}^4.\end{array}$

Next, we consider the case for$N=60$ , Plugging in$N=60$ , and$N_R=\frac{1}{2}N+5=\frac{1}{2}60+5=35$ into eq. (1) we obtain:$\begin{array}{c}{W}_{35}^{60}=\frac{60!}{35!(60−35)!}\\ =5.19×{10}^{16}\end{array}$

To calculate their respective probabilities, we divide$W_{N_R}^R$ by$\underset{=0}{\overset{N}{∑}}{W}_{N_R}^N$ which gives:$P=\frac{W_{N_R}^N}{{\displaystyle \underset{=0}{\overset{N}{∑}}{W}_{N_R}^N.}}=\frac{W_{N_R}^N}{2^N}.$

In the case of$N=20$ , the probability is thus:

$P(N=20)=\frac{1.5×{10}^4}{2^{20}}=0.0143$

In the case of $N=60$, the probability is simply:

$P(N=60)=\frac{5.19×{10}^{16}}{2^{60}}=0.045$

Since$P(N=60)>P(N=20)$ , then a room having$N=60$ particles is more likely to have an imbalance of five particles.

b)

Repeating the same process as in part (a) but replacing $N_R$with $\frac{1}{2}N+0.05N$, for $N=20$, we have $N_R=\frac{1}{2}20+0.05\left(20\right)=11$. So:

$W_{11}^{20}=\frac{20!}{11!(20−11)!}=1.68×{10}^5$

For $N=60$, we have$N_R=\frac{1}{2}60+0.05\left(60\right)=33$

The number of ways is:

$W_{33}^{60}=\frac{60!}{33!(60−33)!}=8.8×{10}^{16}$

Finally, calculating their respective probabilities we have:

$\begin{array}{l}P(N=20)=\frac{1.68×{10}^5}{2^{20}}\\ =\mathrm{0.16.}\end{array}$

and:

$\begin{array}{c}P(N=60)=\frac{8.8×{10}^{16}}{2^{60}}\\ =\mathrm{0.076.}\end{array}$

Since$P(N=60)<P(N=20)$ , then a room having $N=20$particles is more likely to have an imbalance of $5\%$.

(c)

The average number of air molecules in a room is of the order of$>{10}^{23}$ . Thus an imbalance of a trillion air molecules one one side of the room deems insignificant. Because of this, it can still be said that both sides contain precisely half of the total number of air molecules.