Derivation of the probability distribution function of a binomial distribution

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SUMMARY

The discussion focuses on deriving the probability distribution function of a binomial distribution, specifically the formula P(X=r) = ^nC_r p^r q^{n-r}, where X follows a binomial distribution B(n,p). Key components include n, the total number of Bernoulli trials, p, the probability of success, and q, the probability of failure. The derivation is rooted in combinatorial principles, emphasizing the importance of independent trials and the calculation of combinations using factorials.

PREREQUISITES
  • Understanding of binomial distributions and their properties
  • Familiarity with combinatorial mathematics, specifically permutations and combinations
  • Knowledge of Bernoulli trials and their significance in probability theory
  • Basic proficiency in factorial calculations and their applications
NEXT STEPS
  • Study the derivation of the binomial coefficient ^nC_r and its applications
  • Explore the concept of independent trials in probability theory
  • Learn about the Central Limit Theorem and its relation to binomial distributions
  • Investigate applications of binomial distributions in real-world scenarios, such as quality control and risk assessment
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Students and professionals in statistics, mathematicians, and anyone interested in probability theory and its applications in fields such as data science and risk management.

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Is there a way to derive

P (X=r) =^nC_r p^r q^{n-r} , r= 0, 1, 2,..., n

where X: B(n,p)

where n is the total number of bernoulli experiments,

p the probability of success

q, the probability of failure.
 
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Yes, just think about it, it's just simple combinatorics (you omitted to mention independent trials, which is, I'm sure important), and no harder than working out how to choose r from n (in fact it is the same).
 
This is not a problem as matt goes into. Suppose we have XXX and YY,then how many ways can combinations occur? Well there are five elements in 5! ways, but 3 of them are similar and the other two are similar, so it's 5!/(3!2!)=10 distinct ways. YYXXX, YXYXX, YXXYX, YXXXY, XYXXY, XXYXY, XXXYY, XYYXX, XXYYX, XYXYX.
 

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