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If one knows the probability of success $p$,

then how does one calculate the probability of getting success at "first try" using geometric distribution?

Is it simply the probability of success? Are the successive events independent?

mavavilj
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  • Hard to guess about independence if you give us no information at all. – lulu Dec 21 '16 at 11:51
  • @lulu I was just suggesting that if the probability of getting success on first try would be $p$, then why wouldn't the success on any try be also $p$? – mavavilj Dec 21 '16 at 11:53
  • If the random variable has a geometric distribution, then it is the count of trials until the first success in an indefinite sequence of independent Bernoulli trials with an identical success rate. [Sometimes the count of failures before the first success, depending on text.] – Graham Kemp Dec 21 '16 at 11:53
  • Independence depends on context, it's not axiomatic. For sports, say, people sometimes assume "trending" (so that recent victories increase the odds of future victories) or, alternatively, mean reversion (so recent victories lower the probability of subsequent victories). Or you can just have independence. – lulu Dec 21 '16 at 11:56
  • @lulu Well would e.g. winning in lottery be independent? – mavavilj Dec 21 '16 at 11:56
  • If the trials are not independent it is not a geometric distribution. @lulu – Graham Kemp Dec 21 '16 at 11:57
  • In the usual setting, yes. The usual setting is: a random number is uniformly chosen from a fixed list and the gambler selects from the list. – lulu Dec 21 '16 at 11:58
  • @GrahamKemp Oh, agreed. But I think the OP is unsure of the setting (of course I might have the question wrong). – lulu Dec 21 '16 at 12:00

1 Answers1

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If the random variable has a geometric distribution, then it is the count of trials until the first success in an indefinite sequence of independent Bernoulli trials with an identical success rate. [Sometimes the count of failures before the first success, depending on text. We'll assume the former.]

If $X\sim \mathcal{Geo_1}(p)$ then $\mathsf P(X=k)=p(1-p)^{k-1}\quad\bigl[k\in\{1,..,\infty\}\bigr]$.

You appear to require $\mathsf P(X=1)$, the probability that there is one trial until the first success. (The count of trials until the first success is one.)

Graham Kemp
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