A bag contains toys marked 1,2,3,...n.Two tags are chosen at random. 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.
 A: In the first case, the space of possible picks is counted as unordered pairs (or distinct sorted pairs, which might be easier to imagine and easier to write down). So the pick $(1,3)$ and the pick $(3,1)$ are not counted as distinct, and together constitute a single possible pick. Thus, when you subsequently count the number of good picks, you must stay with the same regime.
If you want, you can change this to counting ordered pairs, with a total of $_nC_1\cdot{}_{n-1}C_1$ possible picks and $2n-2$ good picks.
On the other hand, in case 2, the space of possible picks is counted as ordered pairs. So the pick $(1,3)$ and the pick $(3,1)$ are counted as distinct, and together constitute two possible picks. Thus, when you subsequently count the number of good picks, you must stay with the same regime.
It is difficult to change this into an unordered approach, as the possibility of picking the same toy twice makes it non-uniform.
A: 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 $\frac{1}{n}(\frac{1}{n} + \frac{1}{n} + (n-2) \frac{2}{n}) = \frac{2(n-1)}{n^2}$
