A bag contains toys marked 1,2,3,…n.Two tags are chosen at random.
Find P such that the markers will be consecutive integers if the tags are chosen
1)without replacement
2)with replacement
I am having a solution that goes like this:
For case 1:(Without replacement)
There are $nC2$ ways of choosing an element.
Take n=5, you get (1,2)(2,3)(3,4)(4,5) Generalising you can get (n-1) cases of getting consecutive tags.
So the required probability stands $(n-1)/nC2$
For case 2:(With replacement)
There is a sample space of $nC1 * nC1$ or $n^2$.
Here my instructor says that the number of choices will be 2(n-1).
Therefore, the probability of choosing consecutive tags is $\frac{2(n-1)}{n^2}$.
But I don't understand why it's $2*(n-1)$.
Is it simply because we now take both the tags without order for example: 1,2 in two ways (1,2) and (2,1) and they still remain conseceutive.
Best Answer
Here is another way to look at the second part of the question (which is with replacement) -
Let's say you pick the first tag and that is number $k$. If $k$ is $1$ or $n$, there is only possibility to get the consecutive number that we pick $2$ if the first number is $1$ and we pick $(n-1)$ if the first number is $n$. But if $k$ is any of the remaining $(n-2)$ numbers, you have two possibilities for the second number, it is either $(k-1)$ or it is $(k+1)$.
So the probability is $\frac{1}{n}(\frac{1}{n} + \frac{1}{n} + (n-2) \frac{2}{n}) = \frac{2(n-1)}{n^2}$