If X₁, X₂, … , Xk are random variables defined for the same experiment, their joint probability distribution gives the probability of events determined by the collection of random variables: for any collection of sets of numbers {A₁, … , Ak}, the joint probability distribution determines

P( (X₁ is in A₁) and (X₂ is in A₂) and … and (Xk is in Ak) ).

For example, suppose we roll two fair dice independently. Let X₁ be the number of spots that show on the first die, and let X₂ be the total number of spots that show on both dice. Then the joint distribution of X₁ and X₂ is as follows:

P(X₁ = 1, X₂ = 2) = P(X₁ = 1, X₂ = 3) = P(X₁ = 1, X₂ = 4) = P(X₁ = 1, X₂ = 5) = P(X₁ = 1, X₂ = 6) = P(X₁ = 1, X₂ = 7) =

P(X₁ = 2, X₂ = 3) = P(X₁ = 2, X₂ = 4) = P(X₁ = 2, X₂ = 5) = P(X₁ = 2, X₂ = 6) = P(X₁ = 2, X₂ = 7) = P(X₁ = 2, X₂ = 8) = …

… P(X₁ = 6, X₂ = 7) = P(X₁ = 6, X₂ = 8) = P(X₁ = 6, X₂ = 9) = P(X₁ = 6, X₂ = 10) = P(X₁ = 6, X₂ = 11) = P(X₁ = 6, X₂ = 12) = 1/36.

If a collection of random variables is independent, their joint probability distribution is the product of their marginal probability distributions, their individual probability distributions without regard for the value of the other variables. In this example, the marginal probability distribution of X₁ is

P(X₁ = 1) = P(X₁ = 2) = P(X₁ = 3) = P(X₁ = 4) = P(X₁ = 5) = P(X₁ = 6) = 1/6,

and the marginal probability distribution of X₂ is

P(X₂ = 2) = P(X₂ = 12) = 1/36

P(X₂ = 3) = P(X₂ = 11) = 1/18

P(X₂ = 4) = P(X₂ = 10) = 3/36

P(X₂ = 5) = P(X₂ = 9) = 1/9

P(X₂ = 6) = P(X₂ = 8) = 5/36

P(X₂ = 7) = 1/6.

Note that P(X₁ = 1, X₂ = 10) = 0, while P(X₁ = 1)×P(X₂ = 10) = (1/6)(3/36) = 1/72. The joint probability is not equal to the product of the marginal probabilities: X₁ and X₂ are dependent random variables.