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I really understood the explanation of Hypergeometric distribution by looking at this answer but when it comes to Geometric distribution I can't get how they calculate the probability distribution of failures before the first success.

Why do we need to calculate the probability distribution of failures before the first success?Also how did the name geometric came for the distribution?

Community
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justin
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3 Answers3

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2) My guess: it's due to the Geometric series, $\sum_{k=0}^{\infty} x^k$, because $x^n= x \cdot x \ldots x$ n times, this is a product, just like a product of probabilities in the Geometric rv.

1) Trials are $assumed$ independent (in fact iid), hence the outcome of a trial doesn't statistically influence the outcome of the next trial, hence $P(A \cap B) = P(A)P(B)$. Extend it to $n$ identical trials - you get $(1-p)^n$ probability of n failures in $n$ trials. All you need now is to multiply it by the probability of success - $p$.

Alex
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Let there be a sequence of independent experiments all having probability $p$ to succeed. If random variable $X$ is defined by stating that $X=k$ if the $k$-th experiment takes care of the first success then $X$ has geometric distribution with parameter $p$. We could define random variables $X_1,X_2,\dots$ where $X_i=1$ if the $i$-th experiment succeeds and $X_i=0$ if it fails. Then these $X_i$ are iid and have Bernouilli distribution with parameter $p$.

Note that e.g. the event $X=3$ can be described as: $X_1=0\wedge X_2=0\wedge X_3=1$ so that $$P(X=3)=P(X_1=0\wedge X_2=0\wedge X_3=1)=P(X_1=0)P(X_2=0)P(X_3=1)=(1-p)^2p$$

More generally we have: $$P(X=k)=(1-p)^{k-1}p$$

I can't help if it comes to the name geometric. So your question has made me curious as well. I hope someone will react on that.

drhab
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  • :Could you tell me why we are only considering only the first success as stated in your answer? – justin Mar 25 '15 at 11:14
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    That is just a matter of definition (of geometric distribution). More generally you can also state $X=k$ if the $k$-the experiment takes care of the $r$-th success. Then you are dealing with the negative binomial distribution. It can be looked at as a generalization of the geometric distribution (special case $r=1$). – drhab Mar 25 '15 at 11:21
  • :A good explanation of geometric distribution is given here:https://onlinecourses.science.psu.edu/stat414/node/75 .I really understood it.Since I'm not quite efficient to give an answer could you frame the answer or edit your own answer.It might help others too. – justin Mar 26 '15 at 12:06
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We Caclute the probability of failures before first success to arrive at a conclusion about our experiment. E, g. You ask people outside a polling station who voted for whom, and we would wish to find the number of persons who voted for an independent candidate, for that we would ask each and every one, whom you voted for? If n goes on increasing, I, e, yet we didn't found any person who voted for independent candidate even after questioning many voters , obviously p probability of his success declines as per geometric probability distribution graph, so if p is less probability of his loosing will be more as per the formula.(1_p)^n_1 So in order to arrive at a conclusion, we calculate geometric probability of a distribution. I hope the question is answered

Suhail Wani
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