In a set of 5 bottles,1 has a fracture.If you select a pair of bottles, probability that the fractured bottle is chosen is? Its also mentioned that this question is an example of Sampling without replacement
My question is , by the general method of how I do these kind of problems , I would assume that there are two possibilities


*

*You chose a defective bottle and then  a non defective which will have the probability of 
  1/5 X 4/4 = 4/20

    OR


*Choosing a non defective one first and then a defective one which will have the probability of
4/5 x 1/4 =4/20

Total probability is 8/20= 2/5 which doesn't make sense at all, since there is just one bottle to begin with
But the problem is I cant find a fault in my logic. Though I feel both the options are redundant , isn't both two different ways of selecting the pair?. Or is the fact that the phrase "one after the other" is absent a reason on why this might be a wrong approach?
    Thank you in advance

 A: In the general case, say there are $n$ bottles, and you choose $k$ of them. There is only one flawed bottle.
The number of ways to pick $k$ bottles from $n$ is the binomial coefficient $\dbinom{n}{k}=\dfrac{n!}{k!(n-k)!}$.
Now we count the number of these $\dbinom{n}{k}$ arrangements that contain the flawed bottle.
By hand is OK for small $n$ and $k$, for example with $n=4,k=2$ we have $\dbinom{4}{2}=6$ cases:

AB..
A.C.
A..D
BC..
.B.D
..CD

and if the flawed bottle is A (it doesn't actually matter) then there are three cases with an A in them, and so the probability is $\dfrac{3}{6}=0.5$.
More generally, we need to know the number of arrangements that have the flawed bottle.
If we remove the flawed bottle, then we have $n-1$ bottles, and we want to know the number of arrangements of any $k-1$ of them, as we then add the flawed bottle back to give us the total number of arrangements we require. This is $\dbinom{n-1}{k-1}$.
So the general probability is given by:
$$\frac{\dbinom{n-1}{k-1}}{\dbinom{n}{k}}=\frac{(n-1)!}{(k-1)!(n-k)!}\cdot\frac{k!(n-k)!}{n!}=\frac{k}{n}$$
A perhaps more intuitive method is to pick the $k$ bottles first, and then someone assigns a bottle as flawed from the $n$  bottles. You have a $k$ in $n$ chance of getting the flawed bottle.
A: You can think of this problem in two ways. One way is the one you did which is perfectly fine.
The other is to think about the possible outcomes you can have at the end of the two selections. There are $5\choose 2$ total number of selections you can make. Now assume that you have already selected the fractured bottle. There are ${4\choose 1}=4$ different ways of selecting the second bottle. Therefore probability that the fractured bottle is chosen is
$$
\frac{4\choose 1}{5\choose 2}=\frac{4}{10}=\frac{2}{5}
$$
PS: If order of selection is taken into consideration then the total number of outcomes will simply be doubled (since the permutation of the selected pair also matters). But this doesn't affect the final answer as $\frac{{4\choose 1}\times 2}{{5\choose 2}\times 2}=\frac{4\choose 1}{5\choose 2}$
A: Here is another viewpoint.
Suppose you don't know there is a flawed bottle. You choose two of the five bottles. You are then told that one of the five bottles contains a crack. It is equally likely to be any of the five bottles, so each bottle has a probability of $\frac{1}{5}$ of being the cracked one.
You hold two bottles, so the probability of either one of them being the cracked one is $\frac{1}{5}+\frac{1}{5}=\frac{2}{5}$. Note that you can add these two probabilities because they are mutually exclusive events - they cannot both be cracked as there is only one cracked bottle.
Edit: I see now that JMP already briefly mentioned this at the end of his answer.
A: Your logic is perfectly sound. The answer is $\frac{2}{5}$. I think you find it odd that this is more than the probability of choosing one bottle and it being fractured, but this is actually what you must expect. 
Think of the bottles trying to 'clear an exam'. If the exam is easy, more students tend to pass. Thus, each student has a higher chance of passing. You allowing more bottles to 'pass' (choosing more bottles) gives more chance for the fractured bottle to pass too! In the extreme case where all bottles pass, this becomes obvious.
