Prove that you've got a probability density function

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SUMMARY

The discussion focuses on proving that the Bernoulli probability function associated with the scenario of testing cars until two functional ones are found qualifies as a probability density function (PDF). The probability of a car starting is 0.9, while the probability of it not starting is 0.1. The key to proving it as a PDF lies in summing the probabilities of all possible outcomes, specifically using the formula Σ from x=2 to infinity of p^2*(1-p)^(x-2), and incorporating the necessary combinatorial factors to account for the different sequences of successes and failures.

PREREQUISITES
  • Understanding of Bernoulli trials and their properties
  • Familiarity with probability density functions (PDFs)
  • Knowledge of combinatorial mathematics
  • Basic grasp of infinite series and convergence
NEXT STEPS
  • Study the derivation of the Bernoulli distribution and its applications
  • Learn about combinatorial factors in probability calculations
  • Explore the concept of probability density functions in depth
  • Investigate the convergence of infinite series in probability theory
USEFUL FOR

Students in statistics or probability courses, mathematicians focusing on probability theory, and anyone interested in understanding the application of Bernoulli trials in real-world scenarios.

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Homework Statement



Probability of a car starting up is 0.9
Probability of a car NOT starting up is 0.1

Cars are tested until 2 functional cars are found.

Find Bernoulli probability function associated and PROVE that it is a pdf (probability density function).

Homework Equations



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The Attempt at a Solution



To prove that it is a pdf, I think that I need to sum up all of the probabilities associated with the x values and show that it equals 1.

[tex]\Sigma[/tex]x=2 to infinity p^2*(1-p)^(x-2) How do I show that this equals 1? Is it even correct?

Keep in mind that the minimum value for the # of cars tested is 2 because we're looking for 2 functional cars, and once we've got 2, we stop the trials.
 
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It isn't a PDF. You are forgetting a combinatorial factor, aren't you? If you get a second success after x trials and quit, there is more than one way that the first success could have happened. Isn't there?
 
Last edited:

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