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How many births must occur in a family, in order to have one boy and one girl with probability > 80%?

Here is my attempt: If we have n births, all possible arrangements of Bs and Gs (GBGGBBGBG etc) are $2^{n}$. Of those, only one is GGGGG... (only girls) and one is BBBBB... (only boys). All others have both sexes. So in order to find n, we must solve the inequality: $\frac{2^{n} - 2}{2^{n}}\ge 0.80$.

Is this correct? And how do we solve it?

Thank you!

Juan Manuel Prada
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  • Yes, that's correct. Don't need a very large $n$ either...already with $3$ kids the probability of $BBB$ or $GGG$ is $\frac 14$ so you are at $75%$ already. – lulu Oct 02 '20 at 18:57
  • Just for clarity "one boy and one girl" simply means the family two or more children of which there is at least one boy and at least one girl? You don't mean there are exactly one boy and exactly one girl or that two of the children in birth order contain a boy followed immediately by a girl or any strange alternative interpretations? – fleablood Oct 02 '20 at 19:25
  • we want to solve $\frac {2^n -2}{2^n}= 1 - \frac 1{2^{n-1}}\ge 0.8$ or in other words $0.2 \ge \frac 1{2^{n-1}}$. Can you solve that? Assuming $n$ is an natural integer it's just trial and error. – fleablood Oct 02 '20 at 19:31
  • Just to be kooky.... $0.2 \ge \frac 1{2^{n-1}}\implies 1 \ge \frac 5{2^{n-1}}\implies 2^{n-1} \ge 5$ and as $2^3 > 5 > 2^2$ than $n-1 \ge 3$ and $n\ge 4$ and ..... oh, I guess that wasn't as kooky as I thought it'd be. – fleablood Oct 02 '20 at 19:45

1 Answers1

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Yes, that's a good setup. Another way to see this is to take the complement of the probability of having one boy and one girl, which would be the probability of having no boys (or equivalently all girls) or no girls (or equivalently all boys). This probability would be $$\underbrace{\frac{1}{2^n}}_{P(\text{all boys})} + \underbrace{\frac{1}{2^n}}_{P(\text{all girls})} = \frac{2}{2^n}$$

The actual probability would be $1$ minus this value, or equivalently $$1 - \frac{2}{2^n}$$ You are trying to find the minimum value of $n$ such that this probability is greater than $p=0.8$. This inequality is $$1-\frac{2}{2^n} > p \to \frac{2}{2^n} < 1-p$$

To solve this, multiply by $2^n$ on both sides to get $$2 < (1-p) \cdot 2^n$$

Divide by $0.2$ to get $$\frac{2}{1-p} < 2^n$$

Take the log base 2 to get $$\log_2\left(\frac{2}{1-p}\right) = 1 - \log_2(1-p) < n$$

The LHS is approximately $3.3$ for $p = 0.8$, and since $n$ must be an integer, the minimum value of $n$ is $4$.

Varun Vejalla
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