# probability help

Printable View

• Oct 9th 2008, 11:46 AM
alex_2008
probability help
the question is something like this

a group of n girls is given names. there are N different names available. The girls are allotted names at random with 2 or more allowed to have same name. show the probability they will have n different names is (N-1)(N-2)....(N-n+1)/N^(n-1)

i think binomial distribution is involved init somewhere.

please hellllpppppp
• Oct 9th 2008, 12:49 PM
Soroban
Hello, alex_2008!

Quote:

A group of $\displaystyle n$ girls is given names.
There are $\displaystyle N$ different names available.
The girls are allotted names at random with 2 or more allowed to have same name.

Show the probability they will have $\displaystyle n$ different names is: .$\displaystyle \frac{(N-1)(N-2) \cdots (N-n+1)}{N^{n-1}}$

Let's "walk" through this . . .

. . . . $\displaystyle \begin{array}{ccccccc}\text{The 1st person can have any of the }N\text{ na{m}es:} & \qquad\quad Prob &=& \quad\dfrac{N}{N}\quad&=&\quad 1 \\ \\[-3mm] \end{array}$

$\displaystyle \begin{array}{ccccc}\text{The 2nd person can have any of the other }N-1\text{ na{m}es:} & \quad\;\; Prob &=& \dfrac{N-1}{N} \\ \\[-3mm] \end{array}$

$\displaystyle \begin{array}{cccccc}\text{The 3rd person can have any of the other }N-2\text{ na{m}es:} & \quad\;\; Prob &=& \dfrac{N-2}{N} \\ \\[-3mm] \end{array}$

$\displaystyle \begin{array}{ccccc} \vdots & & \vdots \\ \text{The }n^{th}\text{ person can have any of the other }N-n+1\text{ na{m}es:} & Prob &=& \dfrac{N-n+1}{N} \end{array}$

The probability is: . $\displaystyle 1\cdot\underbrace{\frac{N-1}{N}\cdot\frac{N-2}{N} \hdots\frac{N-n+1}{N}}_{n-1\text{ fractions}}$

. . . . . . . . . . . .$\displaystyle = \;\frac{(N-1)(N-2)\cdots(N - n+1)}{N^{n-1}}$

• Oct 9th 2008, 01:18 PM
alex_2008
thanks alot soroban