Discrete Probability Distributions
Modeling countable outcomes in random events using specific probability distributions.
A random variable is a variable whose value is a numerical outcome of a random phenomenon. In A Level Mathematics, we primarily distinguish between two types: discrete and continuous. This topic focuses on discrete random variables, which can only take on a countable number of distinct values, such as the score on a die or the number of defective items in a batch. A discrete probability distribution is a table, graph, or formula that lists all possible values of a discrete random variable and their corresponding probabilities. For any valid distribution, two conditions must hold: P(X=x) ≥ 0 for all possible values x, and the sum of all probabilities must equal 1 (ΣP(X=x) = 1).
We will explore three fundamental discrete distributions used extensively for modeling real-world scenarios.
### The Binomial Distribution
The Binomial distribution is used when an experiment, or 'trial', is repeated a fixed number of times, and we are interested in the total number of 'successes'. For a scenario to be modeled by a Binomial distribution, it must satisfy four key conditions:
If a random variable X follows a Binomial distribution, we write X ~ B(n, p). The probability of achieving exactly r successes in n trials is given by the formula:
P(X = r) = ⁿCᵣ * pʳ * (1-p)ⁿ⁻ʳ
Here, ⁿCᵣ represents the number of combinations in which r successes can occur in n trials. The term pʳ is the probability of r successes, and (1-p)ⁿ⁻ʳ is the probability of the remaining (n-r) trials being failures. The mean, or expected value, of a Binomial distribution is E(X) = np, and the variance is Var(X) = np(1-p).
### The Geometric Distribution
The Geometric distribution is related to the Binomial but addresses a different question. It models the number of trials required to achieve the first success. The conditions are similar to the Binomial, except there is no fixed number of trials:
If a random variable X follows a Geometric distribution, we write X ~ Geo(p). The probability that the first success occurs on the r-th trial is given by the formula:
P(X = r) = (1-p)ʳ⁻¹ * p
This formula represents the probability of having (r-1) consecutive failures, each with probability (1-p), followed by one success with probability p. The mean or expected value of a Geometric distribution, representing the average number of trials needed for the first success, is E(X) = 1/p.
### The Poisson Distribution
The Poisson distribution models the number of events that occur within a fixed interval of time or space. It is appropriate when events happen randomly and independently at a constant average rate. The key conditions are:
If a random variable X follows a Poisson distribution, we write X ~ Po(λ). The probability of observing exactly r events in the interval is given by the formula:
P(X = r) = (e⁻ˡ * λʳ) / r!
Here, 'e' is the base of the natural logarithm. A unique and crucial property of the Poisson distribution is that its mean and variance are equal: E(X) = Var(X) = λ. The rate λ must be adjusted proportionally if the interval changes. For example, if the average is 6 events per hour, it is 3 events per 30 minutes.
#### Poisson Approximation to the Binomial
Under certain conditions, the Poisson distribution can be used as an approximation for the Binomial distribution. This is useful when 'n' is very large and calculations with the Binomial formula become cumbersome. The approximation is valid when:
* n is large (typically n > 50)
* p is small (so that np is moderate, typically np < 5)
In this case, a Binomial distribution B(n, p) can be approximated by a Poisson distribution Po(λ), where the mean is set to be equal: λ = np.
Key Points to Remember
- 1A **discrete random variable** has a countable number of possible values, with ΣP(X=x) = 1.
- 2The **Binomial distribution**, X ~ B(n, p), models the number of successes in a fixed number (n) of independent trials.
- 3The Binomial probability formula is **P(X = r) = ⁿCᵣ * pʳ * (1-p)ⁿ⁻ʳ**, with mean **E(X) = np**.
- 4The **Geometric distribution**, X ~ Geo(p), models the number of trials needed to achieve the first success.
- 5The Geometric probability formula is **P(X = r) = (1-p)ʳ⁻¹ * p**, with mean **E(X) = 1/p**.
- 6The **Poisson distribution**, X ~ Po(λ), models the number of events occurring in a fixed interval of time or space at a constant average rate (λ).
- 7The Poisson probability formula is **P(X = r) = (e⁻ˡ * λʳ) / r!**, where the mean and variance are both equal to **λ**.
- 8A Binomial distribution B(n, p) can be approximated by a Poisson distribution Po(np) when n is large and p is small.
Pakistan Example
Modeling Cricket Outcomes with Binomial Distribution
In Pakistan, cricket is a passion. Consider a star batsman who has a probability of 0.2 of hitting a six on any given delivery he faces that is a 'free hit'. In an upcoming T20 match, it's anticipated he will face 5 'free hit' deliveries. This scenario can be modeled using the **Binomial distribution**, X ~ B(5, 0.2), where n=5 (fixed trials) and p=0.2 (constant probability of success). We can calculate the probability of him hitting exactly 2 sixes on these free hits as: P(X=2) = ⁵C₂ * (0.2)² * (0.8)³ = 10 * 0.04 * 0.512 = 0.2048. This model helps analysts predict player performance and set strategic field placements.
Quick Revision Infographic
Mathematics — Quick Revision
Discrete Probability Distributions
Key Concepts
Formulas to Know
P(X=x) = 1.Binomial probability formula is **P(X = r) = ⁿCᵣ * pʳ * (1-p)ⁿ⁻ʳ**, with mean **E(X) = np**.Geometric probability formula is **P(X = r) = (1-p)ʳ⁻¹ * p**, with mean **E(X) = 1/p**.Poisson probability formula is **P(X = r) = (e⁻ˡ * λʳ) / r!**, where the mean and variance are both equal to **λ**.Modeling Cricket Outcomes with Binomial Distribution
In Pakistan, cricket is a passion. Consider a star batsman who has a probability of 0.2 of hitting a six on any given delivery he faces that is a 'free hit'. In an upcoming T20 match, it's anticipated he will face 5 'free hit' deliveries. This scenario can be modeled using the **Binomial distribution**, X ~ B(5, 0.2), where n=5 (fixed trials) and p=0.2 (constant probability of success). We can calculate the probability of him hitting exactly 2 sixes on these free hits as: P(X=2) = ⁵C₂ * (0.2)² * (0.8)³ = 10 * 0.04 * 0.512 = 0.2048. This model helps analysts predict player performance and set strategic field placements.