I have been trying to derive the binomial distribution through reasoning and logic however I am a bit stuck and in need of possible hints on how to proceed or where I have gone wrong.
My reasoning is as follows:
Let $$n=\text{total number of trials}$$
$$p=\text{probability that a trial ends in success}$$
Then the probability for the case where in n trials there is only one success is
$$p(1-p)^{n-1}$$
And the probability for the case where in n trials there are two successes is
$$p^2(1-p)^{n-2}$$
More generally, the probability that in $n$ trials there are $k$ successes is
$$p^{k}(1-p)^{n-k}$$
As you can see I am far from reaching my goal, any helpful hints or corrections would be appreciated.
Best Answer
Very close! In the probability that in n trials there are k successes is ${n \choose k}p^k (1-p)^{n-k}$ because you need to choose which of the n trials are k successes. $p^k(1-p)^{n-k}$ is the probability of success of a single configuration of k successes in n trials. For example, if you have n=5, k=2, this is the probability of S F F S F (with S=success, F=failure). Another possibility is F S F S F. So you have to multiply your probability with the total number of possibilities, which is $n\choose k$.