Probability Examples

This problem can be solved by first considering the number of different possible combinations of two children. Assuming each child can only be a $\mathrm{Boy}$ or a $\mathrm{Girl}$ (such that we have only two possible outcomes), there are four possible ways to have two children:

Now, we know that one of the children is a $\mathrm{Boy}$, hence the option ($\mathrm{Girl}$, $\mathrm{Girl}$) cannot occur.

This leaves the following possible combinations that have at least one $\mathrm{Boy}$:

Of these three possible combinations, two contain a $\mathrm{Girl}$.

The probability that the second child is a $\mathrm{Girl}$ given the first is a $\mathrm{Boy}$ is then

As of the three possible outcomes above that all contained at least one $\mathrm{Boy}$, only two of the three also contained a $\mathrm{Girl}$.

An easy way to understand and expand this problem is by rephrasing it as bit-strings. Firstly, let

such that the four possibilities become the set of two-bit-string,

The condition that one child is a $\mathrm{Boy}$ states that we want to only consider the bit-strings with at least one 0,

where the superscript $P$ stands for prior.

The condition that the other child is a $\mathrm{Girl}$ then ask, of this reduced set of bit-strings $\mathbb{B}_2^{P}$, how many of the bit-string have a 1:

where the superscript $P$ stands for future.

The probability of this situation occurring is then given by

Using this notation it is easy to see that a question of this form is really just a problem of counting bit-strings.

The problem can now be expanded: if one has ten children, nine of which are a $\mathrm{Boys}$, what is the probability that the tenth is a $\mathrm{Girl}$?

This question can then be rephrased as: of the ten-bit-strings with nine 0s, how many have one 1?

There are eleven ten-bit-strings with nine 0s. Of these eleven, ten have one 1. Hence, the probability of this occurring is given by

One can use the counting bit-string notion to now answer this problem for an arbitrary amount of $\mathrm{Boy}$s or $\mathrm{Girl}$s occurring. Moreover, any binary variable with equal probability of outcome could be treated this way.

One can again use the bit-string notion developed above to answer this problem.

Assuming that the first bit represents the first born child and the second bit represents the second, this question asks: of the two-bit-strings with a 0 as the first bit, how many have a 1 in the second bit?.

We now have

and

From this, the probability can easily be found as

Interestingly, this probability remains the same for an number of children when the order of children is specified. If one lists the prior children in order i.e.,

the some $n$-bit-string $i_p \in \{0,1\}^{\times n}$ has been specified. When now considering the $(n+1)$-bit-strings to model another child, there will always be only two $(n+1)$-bit-strings that have $i_p$ as its first $n$ bits: $i_p0$ and $i_p1$. Hence,

such that probability given this prior will always be $1/2$.