## 6.2 Discrete Probability Distributions

A random variable \(X\) is said to be discrete if it can assume only a finite or countable infinite number of distinct values. A discrete random variable can be defined on both a countable or uncountable sample space. The probability that \(X\) takes on the value \(k\), i.e., \(P(X=k)\), is defined as the sum of the probabilities of all sample points that are assigned the value k. For each value within its domain, we have \(P(X=k) \geq 0\) and that the sum of all probabilities is equal to one. For example, suppose you are flipping a coin three times and associate the outcome “heads” with the number one. The results of this experiment are illustrated in the table below.

Outcome | HHH | HHT | HTH | HTT | THH | THT | TTH | TTT |
---|---|---|---|---|---|---|---|---|

Pr | 1/8 | 1/8 | 1/8 | 1/8 | 1/8 | 1/8 | 1/8 | 1/8 |

X | 3 | 2 | 2 | 1 | 2 | 1 | 1 | 0 |

This experiment can also be represented with a histogram which can be turned into a frequency histogram. This is illustrated for tossing a coin ten times and repeating the experiment 10,000 times.

### 6.2.1 Bernoulli distribution

The Bernoulli (named after Jacob Bernoulli, 1654-1705) distribution is the simplest probability distribution. There are only two outcomes: *Success* and *Failure*. The distribution is characterized by one parameter, i.e., \(p\). The probability mass function is written as \(P(X=1)=p\) and, correspondingly, \(P(X=0)=1-p\). The expected value of the binomial is written as follows:
\[E(x)=1 \cdot p+0 \cdot (1-p)=p\]
To calculate the variance, we have \(E(x^2)=1^2 \cdot p + 0^2 \cdot (1-p)=p\) and \(E(x)^2=p^2\). Thus, we have
\[Var(x)=E(x^2)-E(x)^2=p\cdot(1-p)\]

### 6.2.2 Binomial distribution

The binomial distribution is closely related to the Bernoulli distribution because it represents *repeated* Bernoulli outcomes. The two parameters are \(n\) and \(p\). The number of successes is represented by \(k\). The probability mass function is written as
\[Pr(X=k) = {n \choose k} \cdot p^k \cdot (1-p)^{n-k}\]
The expected value of \(X\) is \(E(X)=n \cdot p\) and the variance is \(Var = n \cdot p \cdot (1-p)\). A situation must meet the following conditions for a random variable \(X\) to have a binomial distribution:

- You have a fixed number of trials involving a random process; let \(n\) be the number of trials.
- You can classify the outcome of each trial into one of two groups: success or failure.
- The probability of success is the same for each trial. Let \(p\) be the probability of success, which means \(1-p\) is the probability of failure.
- The trials are independent, meaning the outcome of one trial does not influence the outcome of any other trial.

Imagine a multiple choice test with three questions and five choices for each question. First, draw a probability tree. What is the probability of two correct responses? Next, the binomial distribution is used to calculate the probability. \[P(X=2) = {3 \choose 2} \cdot 0.2^2 \cdot (1-0.2)^{3-2} =0.096\] and \[P(X=3) = {3 \choose 3} \cdot 0.2^3 \cdot (1-0.2)^{3-3} =0.008\] Summing up both probabilities gives us \(P(X \ge 2) = 0.104\).

The binomial distribution can be used to analyze the issue of overbooking. Assume that an airline as a plane with a seating capacity of 115. The ticket price for each traveler is $400. The airline can overbook the flight, i.e., selling more than 115 tickets, but has to pay $700 in case a person has a valid ticket but needs to be re-booked to another flight. There is a probability of 10% that a booked passenger does not show up. The results for overbooking between 1 and 31 seats are shown in the figure below.

### 6.2.3 Poisson Distribution

The Poisson distribution is used for count data (i.e., \(0,1,2,...\)). The probability mass function for the Poisson distribution is given by the following equation: \[P(X=k)=\frac{\lambda^k \cdot e^{-\lambda}}{k!}\] An example of the Poisson distribution (named after Simeon Denis Poisson, 1781-1840) for different parameter values is shown below.