Twins, Triplets, Quintuplets and Poisson Distributions

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I have a comprehension problem of this exercise

It has been observed in human reproduction that twins occur once in 100 births. If the number of babies in a birth follows a Poisson distribution, calculate the probability of the birth of quintuplets, octuplets.

What would be the \lambda (mean) for compute arrives o quintuplets?

Is it a wrong or incomplete exercise?

Thanks in advance

-OF

Hough D. Young - Statistical treatment of experimental data - Chapter 3 no. 17

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As Andre Nicholas pointed out...the unconditoinal poisson distribution is a bad model for this, as you cannot have a birth of 0 babies. I think its just an ill-conceived (pardon the pun...I couldn't help myself! ;-) attempt to get you to thinking about the poisson.

You are told that the probability of a birth resulting in twins is 0.01. You are also told that the number of births follow a poisson. Therefore, you know that $Poisson(2;\lambda)=.01$ for some $\lambda$. Now, a poisson has the following PMF:

$P(x;\lambda)=\frac{\lambda^x e^{-\lambda}}{x!}$

Entering in the info you were given, we get:

$\frac{\lambda^2 e^{-\lambda}}{2}=.01 \rightarrow \lambda^2 e^{-\lambda}=.02 \rightarrow 2\ln(\lambda)-\lambda=\ln(.02)$

This requires a numerical solution, yielding TWO solutions: $\lambda=0.15$ and $\lambda=8.1$. Since twins are considered relatively rare, its better to consider the case where x=2 is already in the tail of the distribution, as opposed to before the mode. Therefore, you would use 0.15 to get:

$P(X=5)=5.4 x 10^{-7}$

$P(X=8)=5.5 x 10^{-12}$

Overall, I don't think this was a very good homework quesion as its very unrealistic.