Geometric distribution and the probability of getting success at "first try"?

enmobladatn

enmobladatn

Answered question

2022-07-20

Geometric distribution and the probability of getting success at "first try"?
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?

Answer & Explanation

Julianna Bell

Julianna Bell

Beginner2022-07-21Added 19 answers

Step 1
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 G e o 1 ( p ) then P ( X = k ) = p ( 1 p ) k 1 [ k { 1 , . . , } ] .
Step 2
You appear to require P ( X = 1 ), the probability that there is one trial until the first success. (The count of trials until the first success is one.)

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