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Mathematics · Ch 11 — Probability Distributions

Probability Mass Function

11.3.2

Probability Mass Function

Definition 11.3 (Probability mass function). If XX is a discrete random variable taking the values x1,x2,x3,…,xn,…x_1,x_2,x_3,\dots,x_n,\dots, the function f(⋅)f(\cdot) (also written p(⋅)p(\cdot)) defined by

f(xk)=P(X=xk),k=1,2,3,…f(x_k)=P(X=x_k),\qquad k=1,2,3,\dots

is the probability mass function (pmf) of XX.

Theorem 11.1 (characterisation, without proof). f(x)f(x) is a pmf if and only if

(i) f(xk)≥0f(x_k)\ge0 for every kk, and (ii) ∑kf(xk)=1\displaystyle\sum_k f(x_k)=1.

Note

The set of values {f(xk)=P(X=xk)}\{f(x_k)=P(X=x_k)\} is also called the probability distribution of the discrete random variable. It can be presented in three equivalent ways: (a) as a table, (b) as a graph, or (c) as an algebraic expression.

Worked illustrations.

  • Two fair coins, X=X= number of heads: f(0)=14, f(1)=12, f(2)=14f(0)=\tfrac14,\ f(1)=\tfrac12,\ f(2)=\tfrac14 — check: all ≥0\ge0 and 14+12+14=1\tfrac14+\tfrac12+\tfrac14=1. …
Figure 11.4Probability mass function of $f(x)$ for the number of heads when two coins are tossed: $f(0)=\frac{1}{4}$, $f(1)=\frac{1}{2}$, $f(2)=\frac{1}{4}$ (Fig. 11.4)
Fig. 11.4 — Probability mass function of $f(x)$ for the number of heads when two coins are tossed: $f(0)=\frac{1}{4}$, $f(1)=\frac{1}{2}$, $f(2)=\frac{1}{4}$ (Fig. 11.4)

Drawn by us to help you understand the concept clearly, and verified to make sure it's accurate. For exams, practice from your textbook's own diagram.

What this figure shows. Probability mass function of f(x)f(x) for the number of heads when two coins are tossed: f(0)=14f(0)=\frac{1}{4}, f(1)=12f(1)=\frac{1}{2}, $f(2)=\frac{1}{4 …