The binomial experiment

Recall that the multi-stage random experiment is an experiment that is formed by the execution of several other random experiments. In the simplest case, these other random experiments are all identical, independent of each other, and can only have $2$ different outcomes (so called Bernoulli Experiments). If this is the case, we call the multistage experiment a binomial experiment.

The prototypical binomial experiment you should always have in mind is "flipping a coin" a fixed number of times, say $n=5$ times. Indeed, the experiment "flipping a coin once" is a Bernoulli experiment: there are only two outcomes (head or tail), and if executed several times, the results of the experiment are independent of each other other.

Let's be a bit more formal:

Definition 1

Consider an experiment with the two events success ( $S$ ) and failure ( $F$ ). We call this experiment a Bernoulli experiment if $S$ and $F$ form a partition of the sample space, and repeating the experiment does not change the proabability for $S$ and $F$ to occur (they are independent).

In other words: Each time we perform the experiment, either $S$ or $F$ will happen, but not both, and the probabilities $p(S)$ and $p(F)$ do not change and

p(S)+p(F)=1

We call $p(S)$ the success probability and $p(F)$ the failure probability.

Example 1

Flipping a coin with $S$ ="head occurred" and $F$ ="tail occurred" is a Bernoulli experiment (assuming that the case "landing on its edge" is excluded as a possibility). If the coin is fair, the success probability is $p(S)=p(H)=0.5$ and the failure probability is $p(F)=p(T)=0.5$ .
Rolling a die with $S$ ="a six occurred" and $F$ ="no six occurred" is a Bernoulli experiment. If the die is fair, the success probability is $p(S)=1/6$ and the failure probability is $p(F)=5/6$ .

Definition 2

Consider a Bernoulli experiment with success probability $p=p(S)$ , which is repeated $n$ times. This new experiment is called a binomial experiment with repetition number $n$ and success probability $p$ .

Other names for a binomial experiment are a sequence of Bernoulli experiments or a Bernoulli chain of length $n$ and success probability $p$ .

We can make this definition more formal, but it is quite technical. Uncollapse to see.

Show

To be more precise, if we define the events

\begin{array}{lll} S_i&=&\text{"success in repetition $i$"}\\ F_i&=&\text{"failure in repetition $i$"}\end{array}

then the constant (non-changing) probability can be expressed as:

\begin{array}{rll} p&=&p(S_1)=...=p(S_n)\\ 1-p&=&p(F_1)=...=p(F_n)\end{array}

and the independence of the events as:

\begin{array}{rll} p(S_2)&=&p(S_2\vert S_1)\\ &=&p(S_2\vert F_1)\\ p(S_3)&=&p(S_3\vert S_1\cap S_2)\\ &=&p(S_3\vert F_1\cap S_2)\\ &=&p(S_3\vert S_1\cap F_2)\\ &=&p(S_3\vert F_1\cap F_2)\\ ...&&\\ p(S_n)&=& p(S_n \vert S_1\cap ...\cap S_{n-1})\\ &=& p(S_n \vert F_1\cap ...\cap S_{n-1})\\ &...&\\ &=& p(S_n \vert F_1\cap ...\cap F_{n-1})\\ \end{array}

A binomial experiment with parameters $n$ and $p$ is best represented with the following tree:

The tree has $n$ generations, and

\begin{array}{lll} S_i&=&\text{"success in repetition $i$"}\\ F_i&=&\text{"failure in repetition $i$"}\end{array}

and $p$ is the success probability and $1-p$ the failure probability.

Example 2

Are the following experiments binomial experiments? If so, what is the repetition number $n$ and success probability $p$ ?

Flipping a fair coin $10$ times, success is "head occurs".
Rolling a fair die twice, success is "a six occurs".
A box contains $4$ red, $5$ white and $6$ black balls. You select at random $4$ balls. Success is "a black ball is selected".

Solution

yes, $n=10, p=0.5$ .
yes, $n=2, p=1/6$ .
Define $S_1$ ="black ball in first selection" and $p(S_2)$ ="black ball in second selection". If the selection is with replacement then it is a binomial experiment because the repetitions (or events $S_1$ and $S_2$ ) are independent of each other. It is $n=4$ and success probability $p=6/15$ .

However, if the selection is without replacement, then it is not an binomial experiment, because $p(S_1)$ and $p(S_2)$ are not the same: $p(S_1)=6/15$ , and $p(S_2)=6/14$ (first selection was red), $p(S_2)=6/14$ (first selection was white) and $p(S_2)=5/14$ (first selection was black).

Recall the path rules for probability trees. For example, if $n=3$ , the probability for three successes is calculated by multiplying the three branch probabilities:

p(SSS)=p(S_1\cap S_2\cap S_3)=p\cdot p\cdot p=p^3

and the probability that there are two successes followed by a success or a failure is the sum of the corresponding path probabilities:

\begin{array}{lll} p(SSS \text{ or } SSF) &=& \underbrace{p(S_1\cap S_2\cap S_3)}_{\text{path }1}+\underbrace{p(S_1\cap S_2\cap F_3)}_{\text{path }2}\\ &=& \underbrace{p^3}_{\text{path }1}+\underbrace{p^2(1-p)}_{\text{path }2}\end{array}

Example 3

Consider a biased coin with $p(H)=0.2$ and $p(T)=0.8$ . The biased coin is flipped $4$ times.

Is this a binomial experiment? Why?
Draw the corresponding probability tree. Use $H$ and $T$ instead of $S$ and $F$ . The indices can be ignored as well.
Determine the probability $p(HTHT)$ .
Determine the probability that there is exactly one head.

Solution

Yes, experience shows that the head or tail in one repetition does not influence the occurrence of head or tail in the next repetition, and the probability for success (for example head) does also not change.
See figure below.
$p(HTHT)=0.2\cdot 0.8\cdot 0.2\cdot 0.8=\underline{0.0256}$
Add all path probabilities with paths containing exactly one head and three tails:

p(\text{1 head})=4\cdot 0.2\cdot 0.8^3 = \underline{0.4096}

Exercise 1

A fair die is rolled $3$ times. Success is "the die shows a 6". Draw the corresponding probability tree and determine the probability for the event "exactly two times a 6".

Solution

It is a binomial experiment with $n=3$ and $p=\frac{1}{6}$ (success is "a 6 occurs"). The tree is shown below.

The probability of the event $E$ ="exactly two times a $6$ " is the sum of the path probabilities where the paths pass through two $6$ and one "no 6". Thus,

p(E)=3\cdot \frac{1}{6}\cdot\frac{1}{6}\cdot\frac{5}{6}=\underline{0.069}