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I have a nation of population 60 million sharing a finite set of first names ($F$) and last names ($L$). Let's say I have a sample of 3 million people from that nation for which I know first names and last names. From this information I can build a dictionary of first and last names which are a reasonable approximation of $F$ and $L$ and assign a probability to each name and surname (assuming name and surname are independent). So if in my sample I have 2000 Peters and 3000 Griffins, I assign to a random person $X$ probabilities $P$($X$ is Peter) = $2000/3000000$ = $0.00067$ and $P$($X$ is Griffin) = $3000/3000000$ = $0.001$.

My question is, if I have another sample of size $n$ from the same population, how do I calculate the probability that I will encounter in the sample at least two people named Peter Griffin?

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You can assume that $P(f = Peter)$ is independent from $P(l = Griffin)$, so that the probability to find one Peter Griffin is $$P(Y = 1) = P(f = Peter) \times P(l = Griffin)$$ $$P(Y = 0) = P(f = Peter) \times P(l \neq Griffin) + P(f \neq Peter) \times P(l = Griffin) + P(f \neq Peter) \times P(l \neq Griffin)$$

Moreover, you can assume that a sample of $3M$ is representative of your population, and thus the inferred probability could be applied to another sample of size $n$.

To find the probability to find at least two people $(Y >= 2)$ named People Griffin:

$$P(Y >= 2) = 1 - P(Y = 1) - P(Y = 0)$$

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  • $\begingroup$ Sorry, can you help me with a practical R example: $\endgroup$ – Francesco May 11 '15 at 14:11
  • $\begingroup$ Correct me if I'm wrong, but isn't the result: "P(Y = n) = (P(Y = 1) * P(Y = 0))^n" and this "1−P(Y=1)−P(Y=0)" always equals 0? $\endgroup$ – Chris Wohlert Jan 9 '17 at 9:33

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