Probability Distribution - Function, Formula, Table

A probability distribution is a mathematical function or rule that describes how the probabilities of different outcomes are assigned to the possible values of a random variable. It provides a way of modeling the likelihood of each outcome in a random experiment.

While a Frequency Distribution shows how often outcomes occur in a sample or dataset, a probability distribution assigns probabilities to outcomes abstractly, theoretically, regardless of any specific dataset. These probabilities represent the likelihood of each outcome occurring.

Common types of probability distributions include:

Properties of a probability distribution include:

• The probability of each outcome is greater than or equal to zero.
• The sum of the probabilities of all possible outcomes equals 1.

In this article will be covering the key concepts of probability distribution, types of probability distribution, along with the applications in CS.

Probability Distribution of a Random Variable

Now the question comes, how to describe the behavior of a random variable? 

Suppose that our Random Variable only takes finite values, like x₁, x₂, x₃,..., and x_n. i.e., the range of X is the set of n values is {x₁, x₂, x₃,..., and x_n}.
The behavior of X is completely described by giving probabilities for all the values of the random variable X.

The Probability Function of a discrete random variable X is the function p(x) satisfying.

P(x) = P(X = x)

Example: We draw two cards successively with replacement from a well-shuffled deck of 52 cards. Find the probability distribution of finding aces.

Answer: 

Let's define a random variable "X", which means number of aces.
Since we are drawing two cards with replacement from a deck of 52 cards , X can only take on the values 0,1 or 2 as the cards are drawn with replacement, the two draws are independent experiments.

Calculating the probabilities:

P(X = 0) = P(both cards are non-aces)
= P(non-ace) x P(non-ace) 
= \dfrac{48}{52} \times \dfrac{48}{52} = \dfrac {144}{169}

P(X = 1) = P(one of the cards in ace) 
= P(non-ace and then ace) + P(ace and then non-ace)
= P(non-ace) x P(ace) + P(ace) x P(non-ace)
= \dfrac{48}{52} \times \dfrac{4}{52}  + \dfrac{4}{52} \times \dfrac{48}{52} = \dfrac{24}{169}

P(X = 2) = P(Both the cards are aces) 
= P(ace) x P(ace)
= \dfrac{4}{52} \times \dfrac{4}{52} = \dfrac{1}{169}

Now we have the probability distribution for the discrete random variable XXX.
It can be represented in the following table:

X	0	1	2
P(X = x)	144/169	24/169	1/169

It should be noted here that each value of P(X = x) is greater than zero, and the sum of all P(X = x) is equal to 1.

Types of Probability Distributions

We have seen what Probability Distributions are; now we will see different types of Probability Distributions. The Probability Distribution's type is determined by the type of random variable. There are two types of Probability Distributions: 

• Discrete Probability Distributions for Discrete Variables
• Continuous Probability Distribution for Continuous Variables

We will study in detail two types of discrete probability distributions.. 

Discrete Probability Distributions

Discrete Probability Functions applies to discrete random variables, which take countable values (e.g., 0, 1, 2, …). These distributions assign probabilities to individual outcomes.It includes distributions such as Bernoulli, Binomial, and Poisson, which are used to model outcomes that can be counted, as explained below:

Bernoulli Trials

Trials of the random experiment are known as Bernoulli Trials, if they are satisfying below given conditions :

• Finite number of trials are required.
• All trials must be independent. (when the outcome of any trial is independent of the outcome of any other trial.)
• Every trial has two outcomes : success or failure.
• Probability of success remains constant across all trials.

Example: Can throwing a fair die 50 times be considered an example of 50 Bernoulli trials if we define:

• Success is getting an even number (2, 4, or 6),
• Failure as getting an odd number (1, 3, or 5)

Answer:

Yes.,this can be consider as example of 50 Bernoulli trails

• There are 3 even numbers out of 6 possible outcomes, so p = 3/6 = 1 /2
• There are 3 odd numbers out of 6, so q = 3/6 = 1 /2
  So, throwing a fair die 50 times with this definition is a classic example of 50 Bernoulli trials, with p=1/2 and q = 1/2

Binomial Distribution

The binomial distribution models the number of successes (x) in n independent Bernoulli trials, each with success probability p.

For example,

For 1 success in 6 trials, there are 6 possible sequences (e.g., PQQQQQ,QPQQQQ,…PQQQQQ,QPQQQQ,…), each with probability p . (1−p)⁵

Therefore the total Probability is given as = 6. p .(1-p)⁵

Generalizing the idea, if Y is a Binomial Random Variable, the Probability Function P(Y) for the Binomial Distribution for n number of trials is given as:

P(Y) = ⁿC_x p^x(1-p)^n-x 

where

• p is the probability of success in a given trial,
• 'x' be the number of successes, x = 0,1,2...n

Example: When a fair coin is tossed 10 times, find the probability of getting i. exactly six heads. ii. at least six heads.

Answer: 

Every coin tossed can be considered as the Bernoulli trial. Suppose X is the number of heads in this experiment: 
We already know, n = 10, p = 1/2
P(X = x) = ¹⁰C_xp^10-x(1-p)^x  
When x = 6, 
(i) P(x = 6) = ¹⁰C₆ p⁴ (1-p)⁶ = \dfrac{10!}{6!4!}(\dfrac{1}{2})^{6}(\dfrac{1}{2})^{4}\\ \hspace{0.4cm} = \dfrac{7\times8\times9\times10}{2\times3\times4}\times\dfrac{1}{64}\times\dfrac{1}{16} \\ \hspace{0.4cm} = \dfrac{105}{512}

(ii) P(at least 6 heads) = P(X >= 6) = P(X = 6) + P(X=7) + P(X=8)+ P(X=9) + P(X=10) 

= ¹⁰C₆ p⁴ (1 - p)⁶ + ¹⁰C₇ p³ (1 - p)⁷ + ¹⁰C₈ p² (1 - p)⁸ + ¹⁰C₉ p¹(1 - p)⁹ + ¹⁰C₁₀ (1 - p)¹⁰  
=\dfrac{10!}{6!4!}(\dfrac{1}{2})^{10} + \dfrac{10!}{7!3!}(\dfrac{1}{2})^{10} + \dfrac{10!}{8!2!}(\dfrac{1}{2})^{10} + \dfrac{10!}{9!1!}(\dfrac{1}{2})^{10} + \dfrac{10!}{10!}(\dfrac{1}{2})^{10}\\ \hspace{0.5cm} = (\dfrac{10!}{6!4!} + \dfrac{10!}{7!3!}+ \dfrac{10!}{8!2!} + \dfrac{10!}{9!1!}+ \dfrac{10!}{10!})(\dfrac{1}{2})^{10} \\ \hspace{0.5cm} = \dfrac{193}{512}

Negative Binomial Distribution

Negative binomial distribution models the number of trials (n) needed to get k successes, where successes are fixed, but trials vary.

P(X=n)=\frac{n-1}{k−1}p^k(1−p)^{n−k}

Where:

• n = total trials (including the k-th success),
• k = required successes (fixed),
• p = probability of success on a single trial,
• \frac{n-1}{k−1} , the number of ways to arrange (k−1) successes in the first (n−1) trials.

For example,

Probability of getting exactly 3 coupons in 10 pizzas given that probability of success (per pizza): p=0.3

k=3, p = 0.3, n = 10

Therfore, total probability is P(X=10)=(\frac{2}{9})(0.3)^3(0.7)^7≈0.08 (8%)

Poisson Probability Distribution

The Poisson distribution models the number of times an event occurs in a fixed interval of time or space. It is expressed as

f(x; λ) = P(X = x) = (λ^xe^-λ)/x!

where, 

• x is the number of times the event occurred
• e = 2.718...
• λ is the mean value

Example: A bakery sells an average of 5 cupcakes per hour. What’s the probability they sell exactly 3 cupcakes in the next hour?

λ=5 (average rate), k=3 (desired events).

P(X = x) = \frac{e^{-λ}\lambda^k}{x!}\\P(X = 3) = \frac{e^{-5} 5^3}{3!} \approx 0.14

Continuous Probability Distributions

Probability distributions for continuous random variables (uncountable outcomes, e.g., time, height, temperature), such as Uniform and Normal distributions, are explained below.

Uniform Distribution

Uniform Distribution models equally likely outcomes over a closed interval [a,b], where the probability is uniform.

Probability Density Function (PDF) of a Uniform Distribution is given by,

f(x) = \begin{cases} \frac{1}{b - a} & \text{if } a \leq x \leq b, \\ 0 & \text{otherwise.} \end{cases}

Cumulative Distribution Function (CDF) of a Uniform Distribution is given by,

F(x) = \begin{cases} 0 & \text{for } x < a, \\ \frac{x - a}{b - a} & \text{for } x \in [a, b], \\ 1 & \text{for } x > b. \end{cases}

Mean (μ): \mu = \frac{a + b}{2} ​
Variance (σ²): \sigma^2 = \frac{(b - a)^2}{12}

Example:

Random number generator between 0 and 1.

Normal (Gaussian) Distribution

Normal distribution models symmetric, bell-shaped data around a mean (μ) with a spread (σ). It describes data that clusters around a central value, with probabilities decreasing exponentially as values deviate from the mean.

PDF of Normal Distribution is given by,

f(x) = \frac{1}{\sigma \sqrt{2 \pi}} e^{-\frac{(x - \mu)^2}{2 \sigma^2}}

CDF of Normal Distribution is given by,

F(x) = \frac{1}{2} \left[ 1 + \text{erf}\left( \frac{x - \mu}{\sigma \sqrt{2}} \right) \right]

Mean, Median, Mode: μ 
Variance: σ²

Example:

Heights of adults in a population (μ=170, σ=10).

Chi-Square Distribution

The chi-square distribution used in hypothesis testing, especially for goodness-of-fit and independence tests. It only takes non-negative values and is positively skewed.

Degrees of freedom refer to the number of independent values or quantities that can vary in the calculation of a statistic.

• For simple experiments, k = Number of Categories - 1
• In contingency table, k = (Rows - 1) × (Columns - 1) 

Mean: k 
Variance :2k, where k is the degree of freedom

Critical values are used in hypothesis testing to determine whether observed frequencies in a contingency table differ significantly from expected frequencies.

Example,

Observed data; O_i: 55 heads, 45 tails in 100 flips.

Expected (fair coin): E_i: 50 heads, 50 tails.

Null Hypothesis (H₀): The coin is fair (P(Heads)=0.5).
Alternative Hypothesis (H_a​): The coin is biased.

Chi-Square Statistic: \chi^2 = \sum \frac{{(O_i-E_i)}^2}{E_i} = \frac{{(55-50)}^2}{50} +\frac{{(45-50)}^2}{50} = 1.0

Degrees of freedom: k = 2 − 1 = 1.  (since there are 2 categories: heads/tails).

Critical value (α=0.05): {\chi} ^2_{0.95}{(1)}=3.84

Since 1.0 < 3.84, fail to reject H₀ (coin may be fair). The data does not show significant evidence of bias.

Application of Probability Distribution in Computer Science

Probability distributions are used in many areas of computer science are as follows:

• In machine learning, they help make predictions and deal with uncertainty.
• In natural language processing, they are used to model how often words appear.
• In computer vision, they help understand image data and remove noise.
• In networking, distributions like Poisson are used to study how data packets arrive.
• Cryptography uses random numbers based on probability.
• Software testing and reliability also use distributions to predict bugs and failures.

Overall, probability distributions help in building smarter, more reliable, and efficient computer systems.

Solved Questions on Probability Distribution

Question 1: A box contains 4 blue balls and 3 green balls. Find the probability distribution of the number of green balls in a random draw of 3 balls.

Solution:

Given that the total number of balls is 7 out of which 3 have to be drawn at random. On drawing 3 balls the possibilities are all 3 are green, only 2 is green, only 1 is green, and no green. Hence X = 0, 1, 2, 3.

• P(No ball is green) = P(X = 0) = ⁴C₃/⁷C₃ = 4/35
• P(1 ball is green) = P(X = 1) = ³C₁ × ⁴C₂ / ⁷C₃ = 18/35
• P(2 balls are green) = P(X = 2) = ³C₂ × ⁴C₁ / ⁷C₃ = 12/35
• P(All 3 balls are green) = P(X = 3) = ³C₃ / ⁷C₃ = 1/35

Hence, the probability distribution for this problem is given as follows

X	0	1	2	3
P(X)	4/35	18/35	12/35	1/35

Question 2: From a lot of 10 bulbs containing 3 defective ones, 4 bulbs are drawn at random. If X is a random variable that denotes the number of defective bulbs. Find the probability distribution of X.

Solution:

Since, X denotes the number of defective bulbs and there is a maximum of 3 defective bulbs, hence X can take values 0, 1, 2, and 3. Since 4 bulbs are drawn at random, the possible combination of drawing 4 bulbs is given by 10C4.

• P(Getting No defective bulb) = P(X = 0) = ⁷C₄ / ¹⁰C₄ = 1/6
• P(Getting 1 Defective Bulb) = P(X = 1) = ³C₁ × ⁷C₃/¹⁰C₄ = 1/2
• P(Getting 2 defective Bulb) = P(X = 2) = ³C₂ × ⁷C₂/¹⁰C₄ = 3/10
• P(Getting 3 Defective Bulb) = P(X = 3) = ³C₃ × ⁷C₁/¹⁰C₄ = 1/30

Hence Probability Distribution Table is given as follows

X	0	1	2	3
P(X)	1/6	1/2	3/10	1/30

Also Check

• Probability Theory
• Types of Events in Probability
• Probability Density Function
• Events in Probability

Practice Problem Based on Probability Distribution Function

Question 1. A coin is flipped 8 times. What is the probability of getting exactly 5 heads? (Assume the coin is fair.)
Question 2. A dice is rolled until a 4 is rolled. If the first success (rolling a 4) occurs on the 6th roll, how many failures occurred before the success?
Question 3. A customer service center receives an average of 3 calls per hour. What is the probability that they receive exactly 5 calls in an hour?
Question 4. The heights of adult women in a certain population follow a normal distribution with a mean of 64 inches and a standard deviation of 3 inches. What is the probability that a randomly selected woman has a height greater than 66 inches?
Question 5. For a continuous uniform distribution between 2 and 8, find the probability that the random variable is between 4 and 6.
Question 6. A researcher performs a chi-square test to examine if there is a relationship between gender and voting preference in a survey of 150 people. The degrees of freedom for this test are 3. What is the critical value for the chi-square statistic at a 0.05 significance level?
Question 7. A sample of 12 students was taken from a population to test their exam scores. The sample mean is 78, and the sample standard deviation is 5. Test if the sample mean significantly differs from a population mean of 75 at a 0.05 significance level.
Question 8. In a factory, 95% of the machines work well, and 5% are defective. If a machine is randomly selected and found to be defective, what is the probability that it was not properly maintained, given that 20% of the machines are poorly maintained? Use Bayes' Theorem to calculate this.

Answer:-

1. 0.21875.
2. 5
3. 0.1009.
4. 0.2546.
5. 0.3333.
6. 7.81.
7. There is no significant difference between the sample mean and the population mean at the 0.05 significance level.
8. 11.11%.