# Inhabitants probability

• Oct 7th 2011, 05:28 AM
Curtius
Inhabitants probability
Hello,

I have to solve the following example:

"In a town of n+1 inhabitants, a person tells a rumor to a second person, who
in turn repeats it to a third person, etc. At each step the recipient of the rumor is
chosen at random from the n people available.
a) Find the probability that the rumor will be told r times without ever returning
to the originator."

The probability that the second person choose one, that is not the first one, must be (n-2)/(n+1) and also for the 3rd,4th,...,r person. So the final probability should be $\displaystyle \frac{(n-2)^r}{(n+1)^r}$, correct?
• Oct 7th 2011, 08:42 AM
Soroban
Re: Inhabitants probability
Hello,Curtius!

I think you have miscounted.
But then, maybe I did, too . . . Please check my reasoning and my work.

Quote:

$\displaystyle \text{In a town of }n+1\text{ inhabitants, a person tells a rumor to a second person,}$
$\displaystyle \text{who in turn repeats it to a third person, etc. }\;\text{ At each step the recipient}$
$\displaystyle \text{of the rumor is chosen at random from the }n\text{ people available.}$

$\displaystyle \text{a) Find the probability that the rumor will be told }r\text{ times}$
. . . $\displaystyle \text{ without ever returning to the originator.}$

The originator [1] can tell any of the other $\displaystyle n$ people.
. . $\displaystyle P(1) \,=\,\frac{n}{n} \,=\,1$

[2] can tell any of the other $\displaystyle n-1$ people ... excluding [1].
. . $\displaystyle P(2) \,=\,\frac{n-1}{n}$

[3] can tell any of the other $\displaystyle n-1$ people ... excluding [1], but including [2].
. . $\displaystyle P(3) \,=\,\frac{n-1}{n}$

[4] can tell any of the other $\displaystyle n-1$ people ... excluding [1], but including [2] and [3].
. . $\displaystyle P(4) \,=\,\frac{n-1}{n}$
. . . . . . $\displaystyle \vdots$

$\displaystyle [r]$ can tell any of the other $\displaystyle n-1$ people.
. . $\displaystyle P(r) \,=\,\frac{n-1}{n}$

$\displaystyle \text{Prob} \:=\:\underbrace{1\cdot \frac{n-1}{n} \cdot\frac{n-1}{n}\cdot\frac{n-1}{n} \cdots \frac{n-1}{n}}_{r\text{ factors}} \;=\;\left(\frac{n-1}{n}\right)^{r-1}$

• Oct 7th 2011, 09:08 AM
Curtius
Re: Inhabitants probability
Thank you very much. I have also an other question. Please check if it is right.

The second part of the exercise is the following:

Quote:

"b) Find the probability that the rumor will be told r times without ever being
repeated to anybody."
$\displaystyle P(1)=\frac{n}{n}=1$

$\displaystyle P(2)=\frac{n-1}{n}$

$\displaystyle P(3)=\frac{n-2}{n}$

.
.
.

I am not sure what the prob. is for P(r).
• Oct 7th 2011, 11:55 AM
Soroban
Re: Inhabitants probability
Hello again, Curtius!

Quote:

$\displaystyle \text{Find the probability that the rumor will be told }r\text{ times}$
. . $\displaystyle \text{without ever being repeated to anybody.}$

. . $\displaystyle \begin{array}{cccccc}P(1)&=&\dfrac{n}{n}&=&1 \\ \\[-3mm] P(2)&=&\dfrac{n-1}{n} \\ \\[-3mm] P(3) &=& \dfrac{n-2}{n} \\ & \vdots \end{array}$

$\displaystyle \text{I am not sure what the prob. is for }P(r).$

. . $\displaystyle \begin{array}{cccccc}P(1)&=&\dfrac{n}{n}&=&1 \\ \\[-3mm] P(2)&=&\dfrac{n-1}{n} \\ \\[-3mm] P(3) &=& \dfrac{n-2}{n} \\ & \vdots \\ P(r) &=& \boxed{\dfrac{n-(r-1)}{n}} \end{array}$