Sampling with replacement

**slevvio** · November 7th 2007, 01:43 PM

A box contains tickets marked 1,2,...,n. A ticked is drawn at random from the box. Then this ticket is replaced in the box and a second ticket is drawn at random. Find the probabilities of the following events:

a) The first ticket drawn is number 1 and the second ticket is number 2;
b) The numbers on the two tickets are consecutive integers, meaning the first number drawn is one less than the second number drawn.
c) The second number drawn is bigger than the first number drawn.

I can work out part a):

P(first ticket drawn is number 1 $\cap$ 2nd ticket drawn is number 2) = $\frac{1}{n} \times \frac{1}{n} = \frac{1}{n^2}$ but I am having trouble working out b) and c); the answers to which respectively are $\frac{n-1}{n^2}$ and $\frac{1-\frac{1}{n}}{2}$ .

Any help would be appreciated

.

**ThePerfectHacker** · November 7th 2007, 04:35 PM

Originally Posted by slevvio

a) The first ticket drawn is number 1 and the second ticket is number 2;

So we want the first one to be 1 and the second one to be 2. There is only 1 way to do that. In total there are $n$ ways to choose the first card and $n-1$ to choose the second card. So the probability is $\frac{1}{n(n-1)}$

b) The numbers on the two tickets are consecutive integers, meaning the first number drawn is one less than the second number drawn.

Just label all the ways this can happen:
$(1,2),(2,3),...,(n-1,n)$ .
As you can see there are $n-1$ ways of doing this so probability is:
$\frac{n-1}{n(n-1)} = \frac{1}{n}$

c) The second number drawn is bigger than the first number drawn.

Let $p$ be this probability. And let $q$ be the probability of getting a first number larger than the second. So $q$ is the same thing as $p$ except the roles of 1st and 2nd cards are reversed. Hence, $p=q$ . But as you know oppositive probabilities always sum to 1. Since $p,q$ are opposite (meaning anything you chose has to be one of the other) it means $p+q=1$ . That means $p=q=1/2$ .

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