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I am working on this problem of binomial distribution from this book (exercise 3.9)-

The probability of a twin birth is approximately 1/90, and we can assume that an elementary school will have approximately 60 children entering kindergarten (three classes of 20 each). Explain how our "statistically impossible" event can be thought of as the probability of 5 or more successes from a binomial (60, 1/90).

Given solution - The value of n is taken 60 and x ranges from 5 to 60 with p = 1/90.

My approach 1

As far as I understand, here the question is about twins, which occur in pairs. So x = 1 means having 1 twin child which makes no sense. Similarly no odd value of x makes sense. All x should be even, so x should be 10, 12, 14,...., 60. But then the sum of probability distribution would not be equal to 1.

Approach 2

Since twins occur in pairs, and total children are 60, the value of n in binomial distribution should be taken as 30. Then x can range from 5 to 30 (so x = 1 will mean 1 twin pair) and this probability distribution will sum to 1.

Please tell me which approach is correct? Is the book's answer correct or one of my approaches?

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  • $\begingroup$ Seems like the exercise is asking you to calculate the probability "5 or more have a twin", and the assumption is they'd be entering together. So n=60, p=1/90. The author is explicit in giving you the distribution $\endgroup$ – David Peterson Oct 4 '18 at 5:02
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I think what they meant to say is that the probability of a child having been born in a twin birth is 1/90, so on average one out of every 90 children are a twin (or there is a pair of twins in every 180 children). The way they have phrased it, interpreted literally, appears to say that each birth has a probability of 1/90 of producing twins, so the probability of a child being a twin is 2/91.

I doubt the question requires anything more complicated then that, but you could add a condition that if there is a twin then both twins entered kindergarten at the same time. If you wanted to do this the only way I can see to compute this is imagine lining up the children and ask each child if they have a twin, 1-p(=89/90) times they will say no and you move to the next child, the rest of the time they say yes and point out their twin (who you later skip). Each set of answers of yes or no give you a probability, but note that there can be a different number of answers. If you compute the probability of combinations where there are at least 6 twins (or 3 pairs of twins) and the total probability of all valid answer sets you can take the ratio between then and get the probability you are looking for.

I'm not 100% sure this approach is the best and I haven't secretly assumed something more then you've stated but another way (which requires more information) is to use Baye's rule and work out:

$P(N births| M children) = \sum_{N} \frac{P(M Children|N births)\times P(N births)}{P(M children)}$

In which case $P(M children)$ is 1 if M = 60 and $P(M Children|N births)$ can be computed from the probability of twin births.

Then you need to add the probability distribution of the number of births that occured. This suggests to me that I've assumed something in my first approach and that you need more information to answer that question with the condition that all twin pairs attend the same kindergarten.

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