AOS4 Topic 3: Discrete Probability

Introduction

Discrete probability deals with scenarios where the set of possible outcomes of an experiment is countable. This means the outcomes can be listed out, even if the list is infinitely long. It contrasts with continuous probability, where outcomes are part of a continuum.

Random Variables

A random variable is a variable whose value is subject to variations due to chance. In discrete probability, we deal with discrete random variables, which take on countable values. Examples include the number of heads in a series of coin flips, the roll of a die, or the number of defective items in a batch.

Probability Distribution

A discrete probability distribution lists each possible value the random variable can take, along with its probability. For a discrete random variable \(X\), the probability distribution is denoted as \(P(X = x)\), which gives the probability that \(X\) takes the value \(x\).

Conditions for Discrete Probability Distributions

  1. Non-negativity: The probability of each outcome must be between 0 and 1. \[ 0 \leq P(X = x) \leq 1 \]
  2. Normalization: The sum of the probabilities of all possible outcomes must be 1. \[ \sum_{x} P(X = x) = 1 \]

Examples of Discrete Probability Distributions

Bernoulli Distribution

A random variable \(X\) follows a Bernoulli distribution if it has only two possible outcomes: 1 (success) with probability \(p\) and 0 (failure) with probability \(1 - p\).

Probability mass function: \[ P(X = x) = \begin{cases} p & \text{if } x = 1 \\ 1 - p & \text{if } x = 0 \end{cases} \]

Binomial Distribution

A random variable \(X\) follows a binomial distribution if it represents the number of successes in \(n\) independent Bernoulli trials with success probability \(p\).

Probability mass function: \[ P(X = k) = \binom{n}{k} p^k (1 - p)^{n - k} \quad \text{for } k = 0, 1, 2, \ldots, n \] Where \(\binom{n}{k}\) is the binomial coefficient, representing the number of ways to choose \(k\) successes out of \(n\) trials.

Geometric Distribution

A random variable \(X\) follows a geometric distribution if it represents the number of trials needed to get the first success in a series of independent Bernoulli trials with success probability \(p\).

Probability mass function: \[ P(X = k) = (1 - p)^{k - 1} p \quad \text{for } k = 1, 2, 3, \ldots \]

Poisson Distribution

A random variable \(X\) follows a Poisson distribution if it represents the number of events occurring in a fixed interval of time or space, with events occurring independently and at a constant rate \(\lambda\).

Probability mass function: \[ P(X = k) = \frac{\lambda^k e^{-\lambda}}{k!} \quad \text{for } k = 0, 1, 2, \ldots \]

Expectation and Variance

The expected value (or mean) of a discrete random variable \(X\) with probability distribution \(P(X = x)\) is: \[ E(X) = \sum_{x} x P(X = x) \]

The variance of \(X\) is: \[ \text{Var}(X) = \sum_{x} (x - E(X))^2 P(X = x) \]

Alternatively, variance can be calculated using: \[ \text{Var}(X) = E(X^2) - [E(X)]^2 \]

How To Find Discrete Probability Distribution?

A discrete probability distribution can be represented either in the form of a table or with the help of a graph. To find a discrete probability distribution, the probability mass function is required. In other words, to construct a discrete probability distribution, all the values of the discrete random variable and the probabilities associated with them are required. Suppose a fair coin is tossed twice. Say, the discrete probability distribution has to be determined for the number of heads that are observed. The steps are as follows:

Steps:

  1. Determine the sample space of the experiment. When a fair coin is tossed twice the sample space is {HH, HT, TH, TT}. Here, H denotes a head and T represents a tail. Thus, the total number of outcomes is 4.
  2. Define a discrete random variable, X. For the example, let X be the number of heads observed.
  3. Identify the possible values that the variable can assume. There are 3 possible values of X. These are 0 (no head is observed), 1 (exactly one head is observed), and 2 (the coin lands on heads twice).
  4. Calculate the probability associated with each outcome. In the given example, the probability can be calculated by the formula: number of favorable outcomes / total number of possible outcomes.
  5. To get the discrete probability distribution represent the probabilities and the corresponding outcomes in tabular form or in graphical form. This is expressed as follows:

Discrete Probability Distribution Table

x 0 {TT} 1 {HT, TH} 2 {HH}
P(X = x) 1 / 4 = 0.25 2 / 4 = 0.5 1 / 4 = 0.25

Applications of Discrete Probability

Discrete probability is used in various fields such as finance, insurance, quality control, computer science, and many areas of science and engineering. For example:

  • In finance, it is used to model stock prices and risk management.
  • In quality control, it helps in determining the probability of defective products in a batch.
  • In computer science, it is used in algorithms and data structures, particularly in randomized algorithms.

Understanding discrete probability distributions is fundamental to the study of statistics and probability theory, providing the basis for more complex models and analyses.

Example 1

Finding the Value of \( k \) in a Discrete Probability Distribution

Given the following discrete probability distribution:

\( x \) 1 2 3 4
\( P(X = x) \) 0.2 0.5 \( k \) 0.1

Solution:

For a discrete probability distribution, the sum of the probabilities must equal 1:

\[ \sum P(X = x) = 1 \]

Using this property, we set up the equation:

\[ 0.2 + 0.5 + k + 0.1 = 1 \]

Simplifying the equation:

\[ k + 0.8 = 1 \]

\[ k = 1 - 0.8 = 0.2 \]

Answer:

The value of \( k \) is:

\( k = 0.2 \)

Example 2

Finding the Expected Value of a Discrete Probability Distribution

Given the following discrete probability distribution:

\( x \) 0 1 2 3 4
\( P(X = x) \) 0.44 0.36 0.15 0.04 0.01

Solution:

The formula for the expected value of a discrete probability distribution is:

\[ E[X] = \sum x \cdot P(X = x) \]

Substitute the values from the table:

\[ E[X] = (0)(0.44) + (1)(0.36) + (2)(0.15) + (3)(0.04) + (4)(0.01) \]

Perform the calculations:

\[ E[X] = 0 + 0.36 + 0.30 + 0.12 + 0.04 = 0.82 \]

Answer:

The expected value is:

\( E[X] = 0.82 \)

Exercise &&1&& (&&1&& Question)

Is a Normal Distribution a Discrete Probability Distribution?

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Exercise &&2&& (&&1&& Question)

Can the Expected Value of a Discrete Probability Distribution be Negative?

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