Probability Distributions

A probability distribution describes how the values of a random variable are distributed. It tells you which outcomes are likely, which are unlikely, and how probabilities are spread across all possible values.

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Random Variables

A random variable assigns a numerical value to each outcome of a random experiment.

Type	Description	Examples
Discrete	Takes a countable number of values	Number of heads in 10 coin flips, dice rolls
Continuous	Takes any value within an interval	Height, weight, temperature

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Discrete Probability Distributions

Probability Mass Function (PMF)

For a discrete random variable X, the PMF gives the probability of each specific value:

$$P(X = x) = f(x)$$P(X=x)=f(x)

Requirements:

• 0 ≤ f(x) ≤ 1 for all x
• Σ f(x) = 1 (probabilities sum to 1)

Expected Value (Mean)

$$E(X) = \mu = \sum x_{i} \times P(X = x_{i})$$E(X)=μ=∑xi​×P(X=xi​)

Variance

$$\operatorname{Var}(X) = \sigma^{2} = \sum (x_{i} - \mu)^{2} \times P(X = x_{i})$$Var(X)=σ2=∑(xi​−μ)2×P(X=xi​)

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Common Discrete Distributions

Bernoulli Distribution

A single trial with two outcomes (success/failure):

$$\begin{aligned} P(X = 1) &= p && \text{(success)} \\[4pt] P(X = 0) &= 1 - p && \text{(failure)} \end{aligned}$$P(X=1)P(X=0)​=p=1−p​​(success)(failure)​

Mean = p, Variance = p(1 − p)

Binomial Distribution

The number of successes in n independent Bernoulli trials:

$$P(X = k) = C(n, k) \times p^{k} \times (1 - p)^{n - k}$$P(X=k)=C(n,k)×pk×(1−p)n−k

• Mean = np
• Variance = np(1 − p)

Example: Probability of getting exactly 3 heads in 5 coin flips (p = 0.5):

$$P(X = 3) = C(5, 3) \times 0.5^{3} \times 0.5^{2} = 10 \times 0.125 \times 0.25 = 0.3125$$P(X=3)=C(5,3)×0.53×0.52=10×0.125×0.25=0.3125

Poisson Distribution

Models the number of events occurring in a fixed interval of time or space, given a known average rate λ:

$$P(X = k) = \frac{e^{-\lambda} \times \lambda^{k}}{k!}$$P(X=k)=k!e−λ×λk​

• Mean = λ
• Variance = λ

Example: A call centre receives an average of 4 calls per minute. What is the probability of receiving exactly 6 calls in a minute?

$$P(X = 6) = \frac{e^{-4} \times 4^{6}}{6!} \approx 0.1042$$P(X=6)=6!e−4×46​≈0.1042

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Continuous Probability Distributions

Probability Density Function (PDF)

For a continuous random variable, the PDF f(x) describes the relative likelihood of values. Probabilities are found by integrating: