## The Osmond Brothers – a Binomial Example

According to Wikipedia, the Osmonds are an American family music group with a long and varied career – a career that took them from singing barbershop music as children, to achieving success as teen-music idols, to producing a hit television show, and to continued success as solo and group performers (e.g. Donny and Marie Osmond, teen idols in the 1970s, still perform in stage and on television). In addition to their music, one interesting fact about the Osmond family is that the parents, Olive and George Osmond, produced nine children, seven of which are members of the Osmond musical group. What is even more interesting is that the nine children consist of eight boys and one girl (Marie Osmond is the only girl).

The nine children in the Osmond family are: Alan, Wayne, Merrill, Jay, Donny, Marie, Jimmy, George and Tom (the first seven were parts of the family musical group). Assuming that a boy is equally likely as a girl in a pregnancy, the sex of a child is like a coin toss. How likely is it that tossing a coin nine times will result in eight heads and one tail? Intuitively speaking, we know it’s not likely. But “eight heads in nine tosses” can happen (just look at the Osmond family). But how often does this unlikely event happen?

Empirically, we can gather a large sample of families with nine children and see what proportion of such families have eight boys and one girl. Without access to such data, we will use computer to simulate 10,000 coin toss experiments (where each experiment consists of nine tosses), and find out the proportion of experiments with eight heads and one tail. Before we do so, we try a calculation using the binomial distribution. The binomial calculation tells us that it will happen in about 176 times out of 10,000 such experiments, which is similar to the simulated results.

The Binomial Coefficient
Each toss of a coin has two distinct outcomes (head or tail), denoted by H or T. The result of the nine tosses can be described by a string of nine H’s or T’s. There are $2^9=512$ possible strings of H’s and T’s (the string HHTTTHTTH is one example). Because the coin is an unbiased coin, all the 512 strings are equally likely when you toss the coin nine times.

Of the 512 possible strings of H’s and T’s, how many of them have zero head? The answer is one (TTTTTTTT, the string with nine T’s). How many of them have exactly one head? Nine (since the one H can be in any one of the nine positions). How many of them have exactly two heads? There are 36 such strings (HHTTTTTTT, and HTHTTTTTT are two examples). To see this more clearly, we are essentially choosing two positions out of nine positions (two positions for the H’s). There is a combinatorial formula for doing this. The following is the number of ways to choose two objects out of a total of nine objects.

$\displaystyle (1)\ \ \ \ \ \frac{9!}{2! \ (9-2)!}=\frac{9!}{2! \ 7!}=\frac{9 \times 8}{2}=36$

In the above calculation, for any positive integer $k$, $k!$ is called the factorial, which is the product of $k$ and all the positive integers below $k$. For example, $4!=4 \times 3 \times 2 \times 1=24$. In many cases, we do not need to multiply out the factorials. For example, $9!$ divided by $7!$ results in $9 \times 8$.

The notation for the calculation in $(1)$ is $\displaystyle \binom{9}{2}$ or $\displaystyle _9C_2$. The result of the calculation is the number of ways of choosing two objects out of nine objects, or in our problem at hand, the number of 9-character strings consisting of 2 H’s and 7 T’s. For convenience, define $0!=1$. Thus $\displaystyle \binom{9}{9}=1$, which can be obtained by the same idea as in $(1)$, or can be thought of as the number of ways to choose nine objects out of nine objects (only one way).

In general, the number of ways to choose $k$ objects out of $n$ objects is:

$\displaystyle (2)\ \ \ \ \ \binom{n}{k} \text{ or } _nC_k=\frac{n!}{k! \ (n-k)!}$

The number defined in $(2)$ is called the binomial coefficient, so named because the result is the number of ways to separate $n$ objects into two groups (e.g. separating nine positions into H’s and T’s). For more information about this coefficient, see [1] or your favorite introductory statistics textbook.

The following matrix shows the binomial coefficients relevant to the example of tossing a coin nine times (or the Osmond family). Each row in the matrix represents the number of character strings with $k$ H’s and $9-k$ T’s.

$\displaystyle (3)\ \ \ \ \ \begin{pmatrix} \text{k}&\text{ }&\text{Binomial Coefficient} \\\text{ }&\text{ }&\text{ } \\\text{0}&\text{ }&\displaystyle \binom{9}{0}=1 \\\text{2}&\text{ }&\displaystyle \binom{9}{1}=9 \\\text{2}&\text{ }&\displaystyle \binom{9}{2}=36 \\\text{3}&\text{ }&\displaystyle \binom{9}{3}=84 \\\text{4}&\text{ }&\displaystyle \binom{9}{4}=126 \\\text{5}&\text{ }&\displaystyle \binom{9}{5}=126 \\\text{6}&\text{ }&\displaystyle \binom{9}{6}=84 \\\text{7}&\text{ }&\displaystyle \binom{9}{7}=36 \\\text{8}&\text{ }&\displaystyle \binom{9}{8}=9 \\\text{9}&\text{ }&\displaystyle \binom{9}{9}=1 \end{pmatrix}$

The Binomial Probability
In tossing a coin nine times, we want to count the number of heads (consider obtaining a head as a success). Let $X$ be a symbol that denotes the number of heads in tossing a coin nine times. The variable $X$ has 10 possibilities ($0,1,2,3,4,5,6,7,8,9$). The possibility $X=0$ means that there is no head (all nine tosses are tail). The possibility $X=8$ refers to the event there are 8 heads (i.e. the nine tosses have 8 heads and 1 tail).

We would like to compute the ten probabilities for all the possible values of $X$. We use $P(X=k)$ to denote the probability that the value of $X$ is $k$ where $k$ can be any whole number from 0 to 10. So $P(X=\text{8})$ is the probability that there are 8 heads in nine tosses.

As indicated above, there are $2^9=512$ many different strings consisting of H’s and T’s. So the probability $P(X=k)$ is simply the binomial coefficient $\displaystyle \binom{9}{k}$ as a faction of 512. We have:

$\displaystyle (4)\ \ \ \ \ P(X=k)=\binom{n}{k} \times \frac{1}{512}$

The binomial coefficient $\displaystyle \binom{9}{8}=9$. There are only nine strings with 8 H’s and 1 T. Thus the probability of having 8 heads in tossing a fair coin 9 times is $\displaystyle \frac{9}{512}=0.0176$. Thus the outcome of “8 heads and 1 tail” happens about 176 times out of performing 10,000 experiments of tossing the coin 9 times.

The following matrix shows the ten probabilities:

$\displaystyle (5)\ \ \ \ \ \begin{pmatrix} \text{k}&\text{ }&\text{Binomial Coefficient}&\text{P(X=k)} \\\text{ }&\text{ }&\text{ }&\text{ } \\\text{0}&\text{ }&\displaystyle \binom{9}{0}=1&\frac{1}{512}=0.0020 \\\text{2}&\text{ }&\displaystyle \binom{9}{1}=9&\frac{9}{512}=0.0176 \\\text{2}&\text{ }&\displaystyle \binom{9}{2}=36&\displaystyle \frac{36}{512}=0.0703 \\\text{3}&\text{ }&\displaystyle \binom{9}{3}=84&\displaystyle \frac{84}{512}=0.1641 \\\text{4}&\text{ }&\displaystyle \binom{9}{4}=126&\displaystyle \frac{126}{512}=0.2461 \\\text{5}&\text{ }&\displaystyle \binom{9}{5}=126&\displaystyle \frac{126}{512}=0.2461 \\\text{6}&\text{ }&\displaystyle \binom{9}{6}=84&\displaystyle \frac{84}{512}=0.1641 \\\text{7}&\text{ }&\displaystyle \binom{9}{7}=36&\displaystyle \frac{36}{512}=0.0703 \\\text{8}&\text{ }&\displaystyle \binom{9}{8}=9&\displaystyle \frac{9}{512}=0.0176 \\\text{9}&\text{ }&\displaystyle \binom{9}{9}=1&\displaystyle \frac{1}{512}=0.0020 \end{pmatrix}$

Simulating Binomial Probability
Out of 10,000 families with nine children, how many of them are like the Osmond family (with respect to family size and gender make-up of course)? It is hard to say since we do not have empirical data. We do the next best thing by simulation. Using an Excel spreadsheet, we simulate 10,000 experiments of coin tosses (each consisting of nine tosses of a fair coin). The following matrix summarizes the results:

$\displaystyle (6)\ \ \ \ \ \begin{pmatrix} \text{k}&\text{ }&\text{Number of Experiments}&\text{Theoretical Results} \\\text{ }&\text{ }&\text{with k Heads}&\text{based on Binomial Probability } \\\text{ }&\text{ }&\text{ }&\text{ } \\\text{0}&\text{ }&16&20 \\\text{2}&\text{ }&178&176 \\\text{2}&\text{ }&668&703 \\\text{3}&\text{ }&1674&1641 \\\text{4}&\text{ }&2438&2461 \\\text{5}&\text{ }&2521&2461 \\\text{6}&\text{ }&1605&1641 \\\text{7}&\text{ }&698&703 \\\text{8}&\text{ }&174&176 \\\text{9}&\text{ }&28&20 \\\text{ }&\text{ }&\text{ }&\text{ } \\\text{Total}&\text{ }&\text{10,000 }&\text{10,002 } \end{pmatrix}$

The theoretical results in the matrix $(6)$ are obtained by multiplying the probabilities in matrix $(5)$ by 10,000. The middle column in matrix $(6)$ represents the number of simulated experiments with $k$ heads. For example, there are 174 experiments with 8 heads while the theoretical number is 176. Overall, there is remarkable agreement between the simulation and the theoretical binomial distribution. We can conclude that if sex of a child is a coin toss, then the binomial probability describes the gender mix of children in families very well.