Answer:
[tex]P(r)=\frac{(n-3)(n-4)....(n-r)}{(n-2)^{r-2}}[/tex]
Step-by-step explanation:
From the question, we have the following condition:
[tex]p_1=p_2=1,\:and\:p_n=0[/tex]
We know that each student who hears the rumor tells it to a student picked at random from the dormitory (excluding, of course, himself/herself and the person from whom he/she heard the rumor)
The 3rd student can therefore tell the rumour to n-2 students but only n-3 will accept it.
[tex]\implies p(3)=\frac{n-3}{n-2}[/tex]
Consequently, the 4th student must not tell the 1st and second students.
[tex]\implies p(4)=\frac{n-3}{n-2}\times \frac{n-4}{n-2} [/tex]
We can rewrite this to observe a pattern:
[tex]\implies p(4)=\frac{(n-3)(n-4)}{(n-2)^2}[/tex]
[tex]\implies p(4)=\frac{(n-3)(n-4)}{(n-2)^{4-2}}[/tex]
[tex]\implies p(r)=\frac{(n-3)(n-4)(n-5)...(n-r)}{(n-2)^{r-2}}[/tex]
Hence, the probability that the rumor is told r times without coming back to a student who has already is:
[tex]P(r)=\frac{(n-3)(n-4)....(n-r)}{(n-2)^{r-2}}[/tex]
See attachment for complete question
