# Distributions

## Binomial distribution

The **binomial distribution** can be defined as: "the distribution of the number of ‘successes’, X, in a series of *n* independent Bernoulli trials where the probability of success at each trial is *p* and the probability of failure is *q* = 1 − *p*" (Everitt, 2004, p. 40). The Oxford Dictionary of Statistics (Upton & Cook, 2014, p. 43) defines it almost exactly the same. However, there are also those who define it as 'the probability distribution', rather than the absolute count distribution (Zedeck, 2014, p. 28; Porkess, 1991, p. 18).

The definition mentions **Bernoulli trials**, which can be defined as: "a set of *n* independent binary variables in which the *j*th observation is either a ‘success’ or a ‘failure’, with the probability of success, *p*, being the same
for all trials" (Everitt, 2004, p. 35).

The classic example of Bernoulli trials (and for the binomial distribution) is flipping a coin. A coin flip is a binary variable, since there are only two possible outcomes (head or tail, ignoring that strickly speaking it might land on its side). We could define a 'success' as 'head' and if the coin is fair, the probability of success is 0.5. This probability will not change, and remains the same each time we flip the coin.

A Binomial Distribution would then be for example, flipping the coin 5 times and it will then show the probability of having 0, 1, 2, 3, 4, or 5 times a head. The binomial distribution depends on the number of trials (the sample size), usually denoted with **n**, and the probability of success on each trial, usually denoted with **p**.

The distribution will look different, depending on the number of trials and the probability of success.

In figure 1 four examples are shown of the probability distribution of a binomial distribution (the binomial probability mass function (bpmf)).

**Figure 1**

*Four Binomial pmf Examples*

To find probabilities using the binomial distribution, there are three approaches.

**Use some software** (easy)

**Excel**

**Python**

Three different methods to use a binomial distribution with Python:

*With Scipy Library*

Jupyter Notebook: DI - Binomial (scipy).ipynb

*With math Library*

Jupyter Notebook: DI - Binomial (math).ipynb

*Without libraries*

Jupyter Notebook: DI - Binomial (core).ipynb

**R**

R script: DI - Binomial.R

**SPSS**

**Use a binomial distribution table** (old school)

**Click here if you prefer to watch a video on this**

In figure 2 a cumulative binomial distribution table

**Figure 2**

*Cumulative Binomial Distribution Table*

The table shows the probability of **k or less** successes. Lets use an example to illustrate how to work with it.

We have an unfair coin, where the probability of head is 0.8. We throw this coin 5 times and want to know the probability of at least twice a head.

The table columns for p only go up to 0.5, but this is not a problem. The probability of head is 0.8, which means that the probability for tail must be 1 - 0.8 = 0.2. And having at least twice a head, is the same as having at most 3 times a tail. So the probability for at least twice a head = the probability for at most 3 tails.

The number of trials is 5 (we throw the coin 5 times), so we look for the section where n = 5. Then within that section we look for our maximum number of succeses, the 3 (i.e. k = 3).

We finally select the column with our probability of success, the 0.2.

In figure 3 our selection is highlighted.

**Figure 3**

*Cumulative Binomial Distribution Table Example*

The intersection shows 0.993. This is the probability of 3 or less times a tail in five throws, if the chance of a tail is 0.2 each time. It is then also the probability of at least twice a head.

**Use the formulas** (hard core)

The formula for the count of the binomial distribution is given by the binomial coefficient:

\( \binom{n}{k}=\frac{n!}{k!(n-k)!} \)

This formula will tell us how many outcomes there are possible out of *n*, that have exactly *k* successes.

In this formula the \(!\) indicates the factorial operator defined as:

\( x! = \prod_{i=1}^x i = 1\times 2\times 3\times\dots\times\left(x-1\right)\times x\)

While the formula for the binomial probability distribution (the probability mass function) is given by:

\( bpmf(k,n,p)=\binom{n}{k}\times{p^k}\times(1-p)^{n-k} \)

This formula tells us the probability of exactly *k* successes in *n* trials, if the chance of success in each trial is *p*.

The cumulative binomial probability distribution (the cumulative distribution function) the formula is:

\( bcdf(k,n,p)=\sum_{i=0}^{\left\lfloor k\right\rfloor}\binom{n}{i}\times{p^i}\times\left(1-p\right)^{n-i} \)

This formula tells us the probability of *k* or less successes in *n* trials, if the chance of success in each trial is *p*. The brackets around the k in the sum, are the floor function, which means to round k down to the lowest integer. So if k would be 2.99 it would get rounded to 2.

*worked out example*

For example. We have an unfair coin, where the probability of head is 0.8. We throw this coin 5 times and want to know the probability of at least twice a head.

Note that at least twice head + less than twice head should equal 1, so at least twice head = 1 - less than twice head. Less than twice, is the same as 1 or less, and for this we can use the formula, with k=1, n=5 and p=0.8:

\( bcdf(1,5,0.8)=\sum_{i=0}^{\left\lfloor 1\right\rfloor}\binom{5}{i}\times{0.8^i}\times\left(1-0.8\right)^{5-i} \)

\( =\sum_{i=0}^{1}\binom{5}{i}\times{0.8^i}\times(0.2)^{5-i} \)

\( =\binom{5}{0}\times{0.8^0}\times(0.2)^{5-0} + \binom{5}{1}\times{0.8^1}\times\left(0.2\right)^{5-1} \)

\( =\binom{5}{0}\times{1}\times(0.2)^{5} + \binom{5}{1}\times{0.8}\times\left(0.2\right)^{4} \)

\( =\frac{5!}{0!(5-0)!}\times(0.2)^{5} + \frac{5!}{1!(5-1)!}\times{0.8}\times\left(0.2\right)^{4} \)

\( =\frac{5!}{1\times 5!}\times 0.2^{5} + \frac{5!}{1\times 4!}\times 0.8\times 0.2^{4} \)

\( =1\times 0.2^{5} + \frac{5\times4\times3\times2\times1}{4\times3\times2\times1}\times 0.8\times 0.2^{4} \)

\( =0.2^{5} + 5\times 0.8\times 0.2^{4} \)

\( =0.2^{5} + 4\times 0.2^{4} \)

\( =0.00672 \)

Remember that this was to have 1 or less times head, and to get the 'at least twice' we need to subtract this from 1:

1 - 0.00672 = 0.99328

So the probability for at least twice a head out of 5, if the chance for head is 0.8 each throw, is 0.99328.

*Creating the formulas*

If we flip a fair coin 5 times, we have the following possible outcomes:

HHHHH, THHHH, HTHHH, TTHHH, HHTHH, THTHH, HTTHH, TTTHH, HHHTH, THHTH, HTHTH, TTHTH, HHTTH, THTTH, HTTTH, TTTTH, HHHHT, THHHT, HTHHT, TTHHT, HHTHT, THTHT, HTTHT, TTTHT, HHHTT, THHTT, HTHTT, TTHTT, HHTTT, THTTT, HTTTT, TTTTT

For each coin flip we have 2 possible outcomes, we can simply multiply by 2 for each flip we do, so with five flips, we should indeed have 2×2×2×2×2 = 2^{5} = 32 different outcomes. In general we can say that if we have *n*-Bernoulli trials, the total number of possible outcomes is 2^{n}.

How many of those outcomes had exactly 3 'successes'? We define 'success' as 'head'. One of the three can 'choose' to be at any of the 5 positions (trials), the second can then only 'choose' from 4, and the third from only 3. The last two positions are then fixed for the tails. This gives 5×4×3 = 60 possible options. In general we could write this as *n*×(*n*−1)×...×(*n*−*k*+1). Where k is the number of successes we want to know. In mathematics there is a factorial function, which can be defined as:

*n*! = *n*×(*n*−1)×...×(*n*−*k*+1)×(*n*−*k*)×(*n*−*k*−1)×...×2×1

Note that therefor:

(*n*−*k*)! = (*n*−*k*)×(*n*−*k*−1)×...×2×1.

Therefor:

\( \frac{n!}{(n-k)!}=\frac{n\times(n-1)\times...\times(n-k+1)\times(n-k)\times(n-k-1)\times...\times2\times1}{(n-k)\times(n-k-1)\times...\times2\times1} \)

\( =n\times(n-1)\times...\times(n-k+1)\times\frac{(n-k)\times(n-k-1)\times...\times2\times1}{(n-k)\times(n-k-1)\times...\times2\times1} \)

\( =n\times(n-1)\times...\times(n-k+1)\times1 =n\times(n-1)\times...\times(n-k+1) \)

However, if for example the first head 'chooses' position 1, and the second position 2, this would be the same as the first head 'chosing' position 2 and the second position 1. In total we actually have 3×2×1 = 6 times that each option is the same. So we need to divide our result by 6, which gives 60/6 = 10. The 10 options are: TTHHH, THTHH, HTTHH, THHTH, HTHTH, HHTTH, THHHT, HTHHT, HHTHT, HHHTT.

In general the number of options that are the same are *k*×(*k*−1)×...×2×1 = *k*!

This gives for our general formula:

\( \frac{\frac{n!}{(n-k)!}}{k!}=\frac{n!}{k!(n-k)!} \)

This formula is known as the binomial coefficient, and often shortened to the notation as:

\( \binom{n}{k}=\frac{n!}{k!(n-k)!} \)

Lets use this formula to create the binomial distribution of throwing a coin five times:

First if we have exactly zero times head:

\( \binom{5}{0} =\frac{5!}{0!(5-0)!} =\frac{5\times4\times3\times2\times1}{0!\times5!} =\frac{5\times4\times3\times2\times1}{1\times(5\times4\times3\times2\times1)} =1 \)

Note that 0! is defined as 0! = 1. The one option is of course if all five trials are tail: TTTTT.

If we have exactly one time a head:

\( \binom{5}{1} =\frac{5!}{1!(5-1)!} =\frac{5\times4\times3\times2\times1}{1!\times4!} =\frac{5\times4\times3\times2\times1}{1\times(4\times3\times2\times1)} =5 \)

These are: TTTTH, TTTHT, TTHTT, THTTT, HTTTT.

For exactly two times head:

\( \binom{5}{2} =\frac{5!}{2!(5-2)!} =\frac{5\times4\times3\times2\times1}{(2\times1)\times(3\times2\times1)} =\frac{5\times4}{2\times1} =10 \)

The ten options are: TTTHH, TTHTH, THTTH, HTTTH,TTHHT, THTHT, HTTHT, THHTT, HTHTT, HHTTT.

Three times head we used as an example, but here it is again:

\( \binom{5}{3} =\frac{5!}{3!(5-3)!} =\frac{5\times4\times3\times2\times1}{(3\times2\times1)\times(2\times1)} =\frac{5\times4}{2\times1} =10 \)

The ten options are: TTHHH, THTHH, HTTHH, THHTH, HTHTH, HHTTH, THHHT, HTHHT, HHTHT, HHHTT.

Exactly four times head:

\( \binom{5}{4} =\frac{5!}{4!(5-4)!} =\frac{5\times4\times3\times2\times1}{(4\times3\times2\times1)\times1} =\frac{5}{1} =5 \)

The five options are: HHHHT, HHHTH, HHTHH, HTHHH, THHHH.

Exactly five times head:

\( \binom{5}{5} =\frac{5!}{5!(5-5)!} =\frac{5\times4\times3\times2\times1}{(5\times4\times3\times2\times1)\times1} =1 \)

The one option is of course HHHHH.

The binomial distribution for flipping a coin five times is therefor:

Number of successes | Frequency |
---|---|

0 | 1 |

1 | 5 |

2 | 10 |

3 | 10 |

4 | 5 |

5 | 1 |

Now what about those probabilities. If we have an unfair coin, where the probability of head is 0.8, then the probability of tail is 1 − 0.8 = 0.2. We focus again first on the situation where we have three times head. The chances of having three times head is simply 0.8×0.8×0.8 = 0.8^{3} but remember that the other two times then should be tails, so they add 0.2×0.2 = 0.2^{2}. In total we have 0.8^{3}×0.2^{2}=0.02048. This chance we have 10 times, so 10×0.02048 = 0.2048.

If we set p as the probability of success (head in the example), then the chance can be written as:

*p ^{k}*×( 1−

*p*)

*, and then multiplied by the frequency. Altogether:*

^{n-k}\( \binom{n}{k}\times{p^k}\times(1-p)^{n-k} \)

Using this formula we get the following probabilities:

For no heads at all:

\( \binom{5}{0}\times{0.8^0}\times(1-0.8)^{5-0} =1\times{1}\times(0.2)^{5} =0.00032 \)

For exactly one head:

\( \binom{5}{1}\times{0.8^1}\times(1-0.8)^{5-1} =5\times{0.8}\times(0.2)^{4} =0.0064 \)

For exactly two head:

\( \binom{5}{2}\times{0.8^2}\times(1-0.8)^{5-2} =10\times{0.8^2}\times(0.2)^{3} =0.0512 \)

For exactly three head:

\( \binom{5}{3}\times{0.8^3}\times(1-0.8)^{5-3} =10\times{0.8^3}\times(0.2)^{2} =0.2048 \)

For exactly four head:

\( \binom{5}{4}\times{0.8^4}\times(1-0.8)^{5-4} =5\times{0.8^4}\times(0.2)^{1} =0.4096 \)

For exactly five head:

\( \binom{5}{5}\times{0.8^5}\times(1-0.8)^{5-5} =1\times{0.8^5}\times(0.2)^{0} =0.32768 \)

Note that the sum of these should always equal 1, which in the example it does: .00032 + .0064 + .0512 + .2048 + .4096 + .32768 = 1.

To summarise the results we now have:

Number of successes | Frequency | Probability |
---|---|---|

0 | 1 | .00032 |

1 | 5 | .00640 |

2 | 10 | .05120 |

3 | 10 | .20480 |

4 | 5 | .40960 |

5 | 1 | .32768 |

*Loader's Algorithm*

For large values of n, the calculations can take quite long, even for computers. Loader (2002) has derived an algorithm to make this easier. It is shown in figure 4, 5 and 6 as a flowchart.

**Figure 4**

*Loader Algorithm part 1 of 3*

**Figure 5**

*Loader Algorithm part 2 of 3*

**Figure 6**

*Loader Algorithm part 3 of 3*

With large sample sizes the binomial distribution will take quite some time to calculate. Thanks to computers this of course speeds up, but even for a computer this can become a challenge. If you keep on increasing n, the shape of the pmf starts to look like a bell shape, as illustrated in figure 7.

**Figure 7**

*Binomial Distribution Increasing n Example*

As it turns out, if n is large enough, the binomial distribution can be approximated by a normal distribution, which would save time in calculations. Note that if n is large, and p is small, then the binomial distribution could also be approximated with a Poisson distribution

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