Estimation Theory: Get Help Understanding Binomial Proportion

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The discussion focuses on understanding estimation theory, specifically the concept of "binomial proportion." The user seeks clarification on whether the binomial proportion refers to the probability of a successful event, using coin tosses as an example. They express confusion about the term and the related binomial probability formula, which calculates the likelihood of a specific outcome in a random sample. The user provides an example involving voters to illustrate their question about calculating probabilities. Overall, the thread emphasizes the need for clearer explanations of these statistical concepts.
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I got a question on estimation theory. Can anyone explain it to me or give me a link with some tut and solutions so I can get a better understanding.

I got a maths question which i have asked for help but no one has replied yet. Since it was a statistical question I should have posted it here.

The question can be found here:
https://www.physicsforums.com/showthread.php?t=401521

I'm stuck on part (b)
The two issue I think I have with the question are that I don't understand the term "binomial proportion" and estimation theory in general. Can and one explain this to me?
 
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Is the "binomial proportion" just the probability the event is successful?
i.e. for 2 coin toss with 50/ 50 chance of either H or T, the binomial proportion for 2 heads is 1/4?

Is my understanding correct... or is it something else?
 
A binomial probability is the probability that a random sample of size n will have an outcome of x. The equation is

P(x) = n!/(k!-(n-k)!) * pk * qn-k

where k is number of outcomes you want,
p is probability of an outcome,
q is probability of an outcome not happening.

For example, suppose that there are 999 voters in the US. 599 voters were in favor for a certain candidate. If I randomly select 110 voters out of the population, the probability that 56 voters will be in favor of the candidate is

110!/(56!-(110-56)!) * .599599656 * .400400454 = 2.2345492668874894732687678194543e+103 * .599599656 *q54 =
3.632743322467059575061399533725e-13 * q54

gives me a probability of 1.2443365480583675485269518116737 e-34 :confused:
 
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