- #1

WMDhamnekar

MHB

- 378

- 28

Hi,

Each page of a book contains

(b) Show that for large

Conclude that the

My attempt to answer (a)

Now, in our case, n =500 pages and r= 50 misprints. So, there is only 10% chance that any single page has one misprint. ⇒ It is impossible for two or more misprints to be on the same page. So, first condition of Fermi_Dirac statistics is satisfied. All distinguishable arrangements satisfying the first condition have equal probabilities. In our case, It is $\binom{nN}{r}^{-1}$. So, the given answer to (a) is correct.

Now, how to prove (b)?

Each page of a book contains

**N**symbols, possibly misprints. The book contains n =500 pages and r =50 misprints. Show that (a) the probability that pages number 1, 2, . . . , n contain, respectively , $r_1, r_2 , . . . , r_n $ misprints equals $$\frac{\binom{N}{r_1}\binom{N}{r_2}. . .\binom{N}{r_n}}{\binom{nN}{r}};$$(b) Show that for large

**N**, this probability may be approximated by $$\frac{r!}{r_1! r_2! . . . r_n!} n^{-r}$$Conclude that the

**r**misprints are distributed in the**n**pages approximately in accordance with a random distribution of**r**balls in**n**cells.**(Note: The distribution of r misprints among the N available places follows the Fermi-Dirac statistics. Our assertion may be restated as a general limiting property of Fermi-Dirac statistics.)**My attempt to answer (a)

Now, in our case, n =500 pages and r= 50 misprints. So, there is only 10% chance that any single page has one misprint. ⇒ It is impossible for two or more misprints to be on the same page. So, first condition of Fermi_Dirac statistics is satisfied. All distinguishable arrangements satisfying the first condition have equal probabilities. In our case, It is $\binom{nN}{r}^{-1}$. So, the given answer to (a) is correct.

Now, how to prove (b)?

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