# Work out this probability

• Aug 30th 2009, 02:14 AM
DaRush19
Work out this probability
"Write an expression for the probability that in a sequence of r random integers in the range [1,n] no integer occurs more than once"

Let the sequence be represented by $\displaystyle a_{n}= a_{1}, a_{2}, ... a_{r}$ where $\displaystyle r<=n$

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

So $\displaystyle P(a_{1}!=a_{2}!=...!=a_{r}) = (1-\frac{1}{n})^r$
• Aug 30th 2009, 03:24 AM
Plato
Quote:

Originally Posted by DaRush19
"Write an expression for the probability that in a sequence of r random integers in the range [1,n] no integer occurs more than once"

Is my answer right? No it is not.
Let the sequence be represented by $\displaystyle a_{n}= a_{1}, a_{2}, ... a_{r}$ where $\displaystyle r<=n$
$\displaystyle P(a_{1}=a_{2}) = \frac{1}{n}$
$\displaystyle P(a_{1}!=a_{2}) = 1-\frac{1}{n}$
So $\displaystyle P(a_{1}!=a_{2}!=...!=a_{r}) = (1-\frac{1}{n})^r$

Here is a hint.
Roll a die three times. The probability that no number occurs more than once is $\displaystyle \frac{6\cdot 5\cdot 4 }{6^3}$.
• Aug 31st 2009, 07:16 AM
DaRush19
Now that I've worked it out as $\displaystyle (n(n-1)(n-2)..(n-r+1))/n^r$ I'm stuck on when this is approximately equal to $\displaystyle 1/2$

:S