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.

. . $\displaystyle P(r) \,=\,\frac{n-1}{n}$

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

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).

Re: Inhabitants probability

Hello again, Curtius!

Quote:

$\displaystyle \text{Find the probability that the rumor will be told }r\text{ times}$

. . $\displaystyle \text{without ever being repeated to anybody.}$

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

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

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