Calculating Probabilities for Independent Bernoulli Trials

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SUMMARY

The discussion focuses on calculating probabilities for independent Bernoulli trials, specifically addressing the random variable X, which represents the number of failures before the first success with a success probability p. The probability mass function for X is derived as p(x) = p(1-p)^x for x = 0, 1, 2..., although this was later identified as incorrect. The cumulative distribution function for the geometric sequence was also explored, with suggestions that P(X < x) could be expressed as the sum of probabilities P(X = 0) + P(X = 1) + ... + P(X = x-1).

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  • Understanding of Bernoulli trials and their properties
  • Familiarity with probability mass functions (PMFs)
  • Knowledge of cumulative distribution functions (CDFs)
  • Basic concepts of geometric and binomial distributions
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A Question Reads: "Suppose that the random Variable X is the number of failures before the first success in a series of independent Bernoulli trials with success probability p"

a) derive the probability mass function of X
b) what is the probability that X < x where x is a positive integer?

My Answers:

a) this is fairly straight forward. Its just a geometric sequence. p(x) = p(1-p)^x x = 0,1,2...

b) I AM STUCK ON THIS. I know that X < x is just the cumulative distribution function for this geometric sequence, but that just does not work well with this. I tried something different:

"the number of failures until the first sucess? P(X<x) for a geometric sequnce? that means x-1 failures in n trials? (or we could think of it as x-1 sucess in n trials if we look at a failure as a success.)
then would it be X~binomial(n,p)"

Or perhaps it means P(X=0) + P (X=1) + P(X=2) + P(X=3)... + P(X=x-1) for a geometric sequence?

Can somebody help me on it please? thanks!
 
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I just found out now that my a) is wrong :( my prof said its not geometric, but is close to it... anybody have any ideas?
 

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