# Binomial distribution in case of twin children

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?

• 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 – David Peterson Oct 4 '18 at 5:02

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