06/10/2021
Mathematical Insight...
General Principle - "to be able to understand something is to be able to put that into words in a meaningful way"..
Perhaps you have dealt with The Binomial Distribution before, perhaps not. When I got into contact with this model, my first question was - why on earth are Combinations used in the formula for working out probabilities in that model? As a matter of fact they are used to count the number of sequences for a given number of successes.
For example, P(X=2), wants to find the probability of two successes in some experiment of, let's say, five trials.
But why in this instance we use in the formula Combinations? Why do we use 5C2? (In the calculator the button usually is marked "nCr")..
What follows is my attempt at putting an answer into words in an accessible way...
Using our example, we have an experiment of 5 trials and we want to find the probability of 2 successes. P(X=2)= ?
We know that we get a number of sequences which consist of 5 elements. They are p's and q's standing for our probability of success and failure respectively.
And we know that the number of these sequences is given to us by the 5C2 formula.
Let's think about those sequences as positions only which need to be filled in by the probability of the success or the probability of the failure. ..
So, we have:
p1, p2, p3, p4, p5.
This is 5 position sequence to be filled in.
Now in our example we are working out the probability of 2 successes. So instead of having p1 or any other symbol for the position we will have S1 or any other S, depending where in the sequence the success happened.
So for example, if in our experiment we had a success in the first and second trial the sequence would look like below:
S1,S2,p3,p4,p5.
Last three positions would be filled automatically by failures. So we are not concered with them, because they will apear in the sequence only as a consequence of successes in the first and secong position.
Now we come to the main point.
Seeing the positions as distinct objects, we can count in how many ways two of those positions can be chosen. And this counting is being done by the formula for Combinations. We assume that those 2 positions are designated for our 2 success. We are not concerned with other positions because they will be filled automatically by failures. And in that way the 5C2 formula will find all the posible sequences of 5 elements with 2 successes in them...
I hope that this explanation has made this issue more clear now ;-)..