A discrete probability distribution is composed of discrete variables and it elaborates the likelihood of the occurrence of each value of a discrete random variable. If you have studied statistics before, then you may be familiar with many different types of probability distributions, for instance, binomial distribution, normal distribution, and Poisson distribution. We categorize all of these distributions either as a continuous or discrete probability distribution.

 

A random variable whose values can be enumerated is known as a discrete random variable. For example, a set of non-negative integers can be counted, therefore the entire set is a discrete random variable.

 

We match each random variable value with a non-zero probability in a discrete probability distribution. Discrete probability distributions are often given in the form of a table.  The probability distribution is also known as the probability function or the probability mass function.

 

Expected Value Function

We can use the expected value function in discrete probability problems to calculate the mean or expected value. In other words, we can say that the expected value of the function means calculating its average. The expected value function is mathematically denoted as E(x) and has the following formula:

 

E(x) = \sum p_i x_i

 

It means that each value of x is multiplied by its corresponding probability and all the values are added in the end to get the expected value.

 

 

Discrete Probability Distribution Examples

Some of the examples of discrete probability distributions are given below:
  • Friends rolling a dice. If anyone rolls a six, he/she will win.
  • The number of chocolates in a packet.
  • The number of traffic accidents in a city on any given day.
  • The number of students absent from school in five days of the week.
  • The number of customers visiting a restaurant in a month.
  • The number of students graduating from a college in each year.
  • If there is heads up after flipping a coin, then a student will get a chocolate.

 

Rolling a die - Example

For example, if a die is rolled, then the probability of each outcome is represented by the following table:

NumberProbability
11/6
21/6
31/6
41/6
51/6
61/6

 

You can see that in the above table, each value has the same possible outcome. Now, let us calculate the expected value function of the above probability distribution. The formula for the expected value of a function is given below:

 

E(x) = \sum p_i x_i

 

We will multiply \frac{1}{6} with each corresponding value of X and add all the values together to get the expected value function.

 

E(x) = 1 \cdot \frac{1}{6} +  2 \cdot \frac{1}{6} + 3 \cdot \frac{1}{6} + 4 \cdot \frac{1}{6} + 5 \cdot \frac{1}{6} + 6 \cdot \frac{1}{6}

= \frac{21}{6} = 3.5

 

The expected value helps us to determine the average value after the number of trials in an experiment. Of course, the value of the die cannot be 3.5, but we often get a decimal value while computing the expected value function.

 

Representation of the Probability Distribution

The below bar chart represents the possible outcomes of rolling a die.

Bar chart - Discrete Probability Distribution Rolling a die - Bar chart

 

You can see that the probability of all the outcomes is the same.  Now, let us see how to compute the expected value function and values of functions.

 

Example 1

The number of customers visiting a shop on the first five days of the week is given in the tabular form below.

DayNumber of CustomersProbability
1200.20
2100.10
3300.30
4100.10
5300.30

Calculate the expected value function.

Solution

The formula for computing the expected value function is given below:

E(x) = \sum p_i x_i

 

We will multiply the number of customers visiting the shop each day with its corresponding probability. In the end, we will enumerate all the values to get the expected number of customers who can visit the shop on a particular day.

 

E(x) = 10 \cdot 0.20 + 10 \cdot 0.10 + 30 \cdot 0.30 + 10 \cdot 0.10 + 30 \cdot 0.30

E(x) = 24

Hence, an average of 24 customers can visit a shop on a particular day.

 

Example 2

The discrete probability distribution of the variable Z is given in the form of the following table:

ZProbability
260.35
280.10
320.15
350.30
380.10

 

Compute the following values:

1 P (Z \leq 32)

2 P (Z \geq 28)

3 P (Z > 35)

 

Solution

1 P (Z \leq 32)

To compute this value, we will count the probabilities of all the values of Z that are less than and equal to 32. Since we have \leq sign, therefore we will also count the probability of 32 in the final answer.

 

P (Z \leq 32) = 0.15 + 0.10 + 0.35 = 0.60

 

2 P (Z \geq 28)

To compute this value, we will count the probabilities of all the values of Z that are greater than or equal to 28. Since we have \geq sign, therefore we will also count the probability of 28 in the final answer.

P (Z \geq 28) = 0.10 + 0.15 + 0.30 + 0.10= 0.65

 

3 P (Z > 35)

To compute this value, we will count the probabilities of all the values of Z that are greater than 35. Since we only have a greater than symbol, hence we will not count the probability of 35.

P (Z >35) = 0.10

 

 

Example 3

Compute the expected value function or mean from the following table which shows the number of accidents occurring in a city in five days of the week.

DayNumber of accidents
150
275
325
460
540

 

Solution

To calculate the mean value of accidents happening in a city, we need to compute the probability of each day like this:

 

Total number of accidents = 50 + 75  + 25 + 60 + 40 = 250

Probability of Day 1 = \frac{50}{250} = 0.20

Probability of Day 2 = \frac{75}{250} = 0.30

Probability of Day 3 = \frac{25}{250} = 0.10

Probability of Day 4 = \frac{60}{250} = 0.24

Probability of Day 5 = \frac{40}{250} = 0.16

 

After adding all of the above probabilities, the final answer should be 1.

0.20 + 0.30 + 0.10 + 0.24 + 0.16 = 1.00

 

The formula for computing the expected value function is given below:

E(x) = \sum p_i x_i

 

We will multiply each probability with its corresponding value. Finally, we will add all the values together to get the expected value function.

 

E(x) = 0.20 \cdot 50 + 0.30 \cdot 75 + 0.10 \cdot 25 + 0.24 \cdot 60 + 0.16 \cdot 40

E(x) = 10 + 22.5 + 2.5 + 14.4 + 6.4 = 55.8

 

 

Example 4

The discrete probability distribution of the variable X is given in the form of the following table:

XProbability
110.21
130.19
150.46
180.14

 

Compute the following values:

1 P (X \leq 15)

2 P (X \geq 15)

3 P (X > 13)

 

Solution

1 P (X \leq 15)

To compute this value, we will count the probabilities of all the values of X that are less than and equal to 15. Since we have \leq sign, therefore we will also count the probability of 15 in the final answer.

 

P (X \leq 15) = 0.46 + 0.19 + 0.21 = 0.86

 

2 P (X \geq 15)

To compute this value, we will count the probabilities of all the values of X that are greater than and equal to 15. Since we have \geq sign, therefore we will also count the probability of 15 in the final answer.

P (X \geq 15) = 0.46 + 0.14= 0.60

 

3 P (X > 11)

To compute this value, we will count the probabilities of all the values of X that are greater than 11. Since we only have a greater than symbol, hence we will not count the probability of 11.

P (X > 11) = 0.19 + 0.46 + 0.14 = 0.79

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Emma

I am passionate about travelling and currently live and work in Paris. I like to spend my time reading, gardening, running, learning languages and exploring new places.