# Probability Sampling with replacement

• Aug 31st 2009, 08:13 PM
kittykat52688
Probability Sampling with replacement
A box contains tickets marked 1,2,...n. A ticket is drawn at random from tehe box. Then this ticket is replaced in the box and a second ticket is drawn at random. Find the probabilities of the following events:
The first ticket drawn is number 1 and the scond ticket is number 2.
The numbers on the two tickets are consectuive integers, meaning the first number drawn is one less than the second number drawn.
The second number drawn is bigger than the first number drawn.

I understand the first part of it. The first ticket probability is 1/n and second ticket probability is 1/n so it's 1/n^2.
The second one has the first part as 1/n and the second part as (n-1)/n so it's (n-1)/n^2.
I don't understand the third scenario. The answer as shown is (1-1/n)/2?
• Aug 31st 2009, 09:19 PM
CaptainBlack
Quote:

Originally Posted by kittykat52688
A box contains tickets marked 1,2,...n. A ticket is drawn at random from tehe box. Then this ticket is replaced in the box and a second ticket is drawn at random. Find the probabilities of the following events:
The first ticket drawn is number 1 and the scond ticket is number 2.
The numbers on the two tickets are consectuive integers, meaning the first number drawn is one less than the second number drawn.
The second number drawn is bigger than the first number drawn.

I understand the first part of it. The first ticket probability is 1/n and second ticket probability is 1/n so it's 1/n^2.
The second one has the first part as 1/n and the second part as (n-1)/n so it's (n-1)/n^2.
I don't understand the third scenario. The answer as shown is (1-1/n)/2?

You have the second part right but the explanation is wrong.

The first ticket $k$ must be one of $1, 2,\ ..\ n-1$ (which occurs with prob $(n-1)/n$ and the second must be $k+1$ (drawn from $n$ tickets) which occurs (given that the first ticket is chosen correctly) with probability $1/n$. Hence the required probability is $(n-1)/n^2$

CB
• Aug 31st 2009, 09:24 PM
CaptainBlack
Quote:

Originally Posted by kittykat52688
A box contains tickets marked 1,2,...n. A ticket is drawn at random from tehe box. Then this ticket is replaced in the box and a second ticket is drawn at random. Find the probabilities of the following events:
The first ticket drawn is number 1 and the scond ticket is number 2.
The numbers on the two tickets are consectuive integers, meaning the first number drawn is one less than the second number drawn.
The second number drawn is bigger than the first number drawn.

I understand the first part of it. The first ticket probability is 1/n and second ticket probability is 1/n so it's 1/n^2.
The second one has the first part as 1/n and the second part as (n-1)/n so it's (n-1)/n^2.
I don't understand the third scenario. The answer as shown is (1-1/n)/2?

For the third part we can consider three possibilities, the first is greater than the second, the second is greater than the first, and that they are equal.

The first two are obviously equal (by symmetry), and the second is $1/n$.

These sum to one so let $p$ denote the probability that the second is greater than the first, then we have:

$2p+1/n=1$

hence:

$p=(1-1/n)/2$

CB
• Sep 18th 2012, 12:39 PM
mariajose621
Re: Probability Sampling with replacement
Hi, I have a similar question.
A random sample n = 3 is selected from N with replacement.
I do not understand why the probability of having , two distinct
units is P = 3*(N-1)/(N^2), and three distinct units is P = (N -1)(N -2)/N^2

Thanks