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sherrellbc
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Homework Statement
The binomial coefficient formula can be used in many applications. What I have been learning about is its application in Binomial Distributions.
Define a random variable X to be the number of heads associated with each flip.
Imagine you have, for example, 5 coins. What is the probability that X = 2?
That is, P(X=2)
So, below I have prove how you can use the nCk formula for situations like this one in which you have 1 of two possible outcomes. But what if you were to have many more? Imagine instead of H/T, you have A,B,C,D. How can we apply the nCk in situations like this?
Homework Equations
## {(n k)} = {n!}/{k!(n-k)!} ##
The Attempt at a Solution
Without applying the formula, it is easy to derive(essentially the formula in the process) the solution to the problem.
Given 5 "spaces", the first H has 5 potential slots, then for the next H there remains 4. That is,
_ _ _ _ _
5 4
Which is 5*4 and can be written as, 5!/(5-2)!
Then we want to divide out by all possible permutations of 2 "things", in this case heads.
So, we have:
## {(5 2)} = {5!}/{2!(5-2)!} = 10*{1/2^6}##
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But, what if we had more possible outcomes? (i.e. A, B, C ... )
The use of nCk in the previous manner worked because we "knew" what the other spaces would hold. That is,
_ _ _ _ _
5 4
We know that the empty spaces here WOULD be tails, since we defined Heads as success.
For example:
H H T T T
H T H H H
H T T H H
H T T T H
...
What if we used the same logic in the case where we have many more possible outcomes?(1, B, C ... )
Redefine X to be: Number of A's
Find P(X=2)
Using the same logic, we resolve the same solution! That:
## {(5 2)} = {5!}/{2!(5-2)!} = 10##
That only difference being that we have no information on probability of A, B, C happening as we did with the Heads/Tails flips.
So, we are essentially saying here that we have 10 UNIQUE combinations of two A's with any other selection of B, C, D .. ect.
How is this possible?
tl;dr, same result with more possible outcomes does not make immediate sense.