Probability - Not a clue!

ShaShaBear

So, Im back again with yet more statistics I don't understand (Im not sure if this is a regular thing or if its just me struggling to grasp the basics). This time it's probability.
In the exam, I am given a formula sheet with the info I need for probability and probability distribution. The problem is, I am having issues pairing questions with formulas.
The question I am looking at is as follows:
A QC analyst inspects a random sample of four PDAs from a shipment of twenty to ensure there are no faulty devices. What is the probability that she will find exactly two faulty devices in her sample:
If a shipment contains 4 faulty devices
If a shipment contains 8 faulty devices

After a lot of brain melt, I decided I needed the formula for a hypergeometric distribution, which is given to me as follows in the exam:
P(xln, X, N) =
( X ) * (N-X)
( x ) (n-x)
___________
(N)
(n)
(The double brackets are meant to indicate a large continuous bracket)

I know what each symbol signifies in this question, I just have no idea what they expect me to do in using this formula. Anyone care to shed some light on how I use the formula to get an answer?

sponsoredwalk

A nice way to look at all probability questions in a way that, to me, helps you locate when & where combinatorics is needed, is by following this procedure:

Random experiment: Selecting 4 PDA's From a shipment of 20
Possible Outcomes: (Good, Good, Good, Bad), (Good, Bad, Good, Good),... (this step is just to get a feel for the sample space)
Sample Space: "Set of All Possible Ways of Selecting 4 Elements From a Set of 20 Possibilities (Where Order Doesn't Matter)"
Size of Sample Space: C(20,4) = 20!/4!(20-4)!
Event A: "Select Exactly Two Faulty Devices"
This is Composed of Selecting 2 Faulty Devices From a Possibility of 4 (where order doesn't matter) then Selecting 2 Good Devices From the remaining possibility of 20 - 4 = 16 (again where order doesn't matter)
Size of Event A: C(4,2)·C(16,2) (this follows from the multiplication principle, if one thing can be done m ways & another n ways then both together can be done in a total of mn ways - & this itself is nothing but a tree diagram formalized)
Probability of A Occurring: |A|/|S| = C(4,2)·C(16,2)/C(20,4)

See if you can do the above when there are 8 faulty devices.

This process of analyzing things is nothing more than the hypergeometric distribution expanded. The way we inject the hypergeometric distribution into this process is through the concept of a random variable & it's distribution function.

Random Variable X: "Number of faulty devices selected"
X(good, good,bad,good) = 1
X(good,bad,bad,good) = 2
Basically X is a function from Sample Space S to R that plugs in possible outcomes & tells you the number of faulty devices.
Distribution Function of Random Variable: F(X = x) = P{X = x}
This is where we interject hypergeometricity... The question has stated that X = 2, i.e. the number of faulty devices is 2, thus F(X = 2) = P{X = 2} = |A|/|S| = C(4,2)·C(16,2)/C(20,4)
So we can use the probability calculation I did above or we can recognize this a process with two outcomes (success or fail, where success or fail is either faulty of not-faulty) & that there is sampling without replacement thus making this a hypergeometric process instead of a binomial process.
Therefore P{X = 2} = f(2,20,4,4) (taken the notation from the urn example in that link) = "hypergeometric formula which gives exact same value as C(4,2)·C(16,2)/C(20,4)". Notice the formula is saying exactly, exactly, the same thing that I said in steps above, it's nothing but a shortcut in this case...

Again see if you can do all of this when there are 8 faulty devices...

Literally the only book that made sense to me in showing me how to deal with things this way was this book (which the examples in sections 1.2, 1.3, 2.1 & 2.2 will offer some perfect opportunities to quickly practice examples in the above systematic way!) so you might appreciate checking that out.