What is Geometric distribution: Definition and 29 Discussions

In probability theory and statistics, the geometric distribution is either one of two discrete probability distributions:

The probability distribution of the number X of Bernoulli trials needed to get one success, supported on the set { 1, 2, 3, ... }
The probability distribution of the number Y = X − 1 of failures before the first success, supported on the set { 0, 1, 2, 3, ... }Which of these one calls "the" geometric distribution is a matter of convention and convenience.
These two different geometric distributions should not be confused with each other. Often, the name shifted geometric distribution is adopted for the former one (distribution of the number X); however, to avoid ambiguity, it is considered wise to indicate which is intended, by mentioning the support explicitly.
The geometric distribution gives the probability that the first occurrence of success requires k independent trials, each with success probability p. If the probability of success on each trial is p, then the probability that the kth trial (out of k trials) is the first success is




Pr
(
X
=
k
)
=
(
1

p

)

k

1


p


{\displaystyle \Pr(X=k)=(1-p)^{k-1}p}
for k = 1, 2, 3, ....
The above form of the geometric distribution is used for modeling the number of trials up to and including the first success. By contrast, the following form of the geometric distribution is used for modeling the number of failures until the first success:




Pr
(
Y
=
k
)
=
Pr
(
X
=
k
+
1
)
=
(
1

p

)

k


p


{\displaystyle \Pr(Y=k)=\Pr(X=k+1)=(1-p)^{k}p}
for k = 0, 1, 2, 3, ....
In either case, the sequence of probabilities is a geometric sequence.
For example, suppose an ordinary die is thrown repeatedly until the first time a "1" appears. The probability distribution of the number of times it is thrown is supported on the infinite set { 1, 2, 3, ... } and is a geometric distribution with p = 1/6.
The geometric distribution is denoted by Geo(p) where 0 < p ≤ 1.

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  1. K

    I Geometric Distribution Problem Clarification

    (Geometric). The probability of being seriously injured in a car crash in an unspecified location is about .1% per hour. A driver is required to traverse this area for 1200 hours in the course of a year. What is the probability that the driver will be seriously injured during the course of the...
  2. U

    Geometric Distribution: Finding Specific p Value for Mean Calculation

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  3. christang_1023

    I How to understand this property of Geometric Distribution

    There is a property to geometric distribution, $$\text{Geometric distribution } Pr(x=n+k|x>n)=P(k)$$. I understand it in such a way: ##X## is independent, that's to say after there are ##(n+k-1)## successive failures, ##k## additional trials performed afterward won't be impacted, so these ##k##...
  4. binbagsss

    Moments from characteristic function geometric distribution

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  5. Alex_Doge

    Proof that an interval is a confidence interval for Geom(q)

    Hello Physicsforum Homework Statement I have a problem proving this: Given C(x)=[0, 3/x] for all x\in\chi, with \chi=\Omega being the sample space and P_q=Geom(q) being the geometric distribution. I have to show that C(x) is a confidence Interval for q but I don't know how to get started...
  6. W

    Markov's Inequality for Geometric Distribution.

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  8. throneoo

    Geometric distribution Problem

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  9. S

    How to solve a geometric distribution problem with a biased coin?

    A boy is playing with a biased coin. The probabilty of obtaining a head with the coin is 0.4. Determine the probability that the boy will require at least eleven tosses before obtaining his third head. I have been trying but can't get it at all... Can someone please explain me how to solve...
  10. H

    Geometric Distribution Probability problem

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  11. M

    Fitting a geometric distribution to data

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  12. M

    Discrete Random Variables - Geometric Distribution

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  13. T

    Geometric Distribution

    Question: Two faulty machines, M1 and M2, are repeatedly run synchronously in parallel (i.e., both machines execute one run, then both execute a second run, and so on). On each run, M1 fails with probability p1 and M2 with probability p2, all failure events being independent. Let the random...
  14. T

    Geometric Distribution

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  15. M

    Median geometric distribution

    Homework Statement How do you find the median of the geometric distribution? Homework Equations M is median if P(X>=M) >= 1/2 and P(X<=M)>=1/2. The Attempt at a Solution I have found this inequality using the geometric series: (m-1)*log(1-p) >= 1/2
  16. S

    Computing the Mean of a Geometric Distribution

    Homework Statement Problem H-10. We will compute the mean of the geometric distribution. (Note: It's also possible to compute E(X^2) and then Var(X) = E(X^2)−(E(X))^2 by steps similar to these.) (a) Show that E(X) = (k=1 to infinity summation symbol) (k *q^k−1* p) where q = 1−p. (b)...
  17. D

    Geometric Distribution Coin Flip

    Consider the following experiment: a coin that lands heads with probability p is flipped once; if on this first flip it came up H, it is then repeatedly flipped until a T occurs; else, if on the first...
  18. T

    Statistics: geometric distribution proof problem

    Statistics: geometric distribution "proof" problem Homework Statement If Y has a geometric distribution with success probability p, show that: P(Y = an odd integer) = \frac{p}{1-q^{2}} Homework Equations p(y)=p(q)^{2} The Attempt at a Solution p(1)=pq^0 p(3)=pq^2 p(5)=pq^4...
  19. J

    Geometric Distribution Question

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  20. F

    Geometric Distribution problem

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  21. D

    Geometric distribution problem

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  22. C

    Sufficient Estimator for a Geometric Distribution

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  23. B

    Estimating Geometric Distribution

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  24. R

    Understanding Geometric Distribution

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  25. 6

    Geometric Distribution and probability

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  26. M

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  27. S

    Geometric distribution problem

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  28. I

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  29. C

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