MHB Probability of misprints on each page

[WMDhamnekar]

Hi,
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)?

 

[pbuk]

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.

This is nonsense. Of course it is possible for two or more misprints to be on the same page, what is not possible is for two misprints to be in the same symbol. If you are still interested in answering this question I suggest you start again at the beginning.

 

[WMDhamnekar]

My answer to (a) Yes. N= Total symbols on each page. r= 50 misprints. So, what is impossible is two or more misprints occurring in the same symbol. 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, we showed that the probability the pages number 1,2,... ,500 contain, respectively, $r_1, r_2, ...,r_{500}$ misprints equals $\displaystyle\frac{\binom{N}{r_1} \binom{N}{r_2} ... \binom{N}{r_n}}{\binom{nN}{r}}$ because the probability that page number 1,2 ,..., 500 contain $r_1 , r_2, ..., r_{500}$ misprints, respectively, (where $r_1 + r_2 + ... + r_{500} =50$) equals $\binom{500N}{50}^{-1}$ under Fermi-Dirac statistics provided each r_j equals 0 or 1.

My answer to (b): We know that r₁ + r₂ + ... + r₅₀₀= 50. The number of placements of 50 misprints on 500 pages resulting in the occupancy numbers r₁ ,r₂ , ... ,r₅₀₀ is given by $\binom{500}{50}$

Assuming that all 500⁵⁰ possible placements are equally probable, the probability to obtain the given occupancy numbers r₁, ... , r₅₀₀ equals $\displaystyle\frac{50!}{r_1!r_2!...r_{500}!}$500^-50

 

[PeroK]

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.

I don't agree wiht that.

 

[malawi_glenn]

What is $50_{500}!$?

 

[WMDhamnekar]

What is 50₅₀₀!

50₅₀₀= r₅₀₀ where r_k stands for number of misprints in the kth page. r= 50 misprints. n=500 pages

 

[malawi_glenn]

@WMDhamnekar why should a number have an index?

 

[WMDhamnekar]

@WMDhamnekar why should a number have an index?

I edited my post #3. Please Recheck it.