Editing Independence (probability theory) (section)

==Examples==
===Rolling dice===
The event of getting a 6 the first time a die is rolled and the event of getting a 6 the second time are ''independent''. By contrast, the event of getting a 6 the first time a die is rolled and the event that the sum of the numbers seen on the first and second trial is 8 are ''not'' independent.

===Drawing cards===
If two cards are drawn ''with'' replacement from a deck of cards, the event of drawing a red card on the first trial and that of drawing a red card on the second trial are ''independent''. By contrast, if two cards are drawn ''without'' replacement from a deck of cards, the event of drawing a red card on the first trial and that of drawing a red card on the second trial are ''not'' independent, because a deck that has had a red card removed has proportionately fewer red cards.

===Pairwise and mutual independence===
[[File:Pairwise independent.svg|thumb|Pairwise independent, but not mutually independent, events]]
[[File:Mutually independent.svg|thumb|Mutually independent events]]

Consider the two probability spaces shown. In both cases, <math>\mathrm{P}(A) = \mathrm{P}(B) = 1/2</math> and <math>\mathrm{P}(C) = 1/4</math>. The events in the first space are pairwise independent because <math>\mathrm{P}(A|B) = \mathrm{P}(A|C)=1/2=\mathrm{P}(A)</math>, <math>\mathrm{P}(B|A) = \mathrm{P}(B|C)=1/2=\mathrm{P}(B)</math>, and <math>\mathrm{P}(C|A) = \mathrm{P}(C|B)=1/4=\mathrm{P}(C)</math>; but the three events are not mutually independent. The events in the second space are both pairwise independent and mutually independent. To illustrate the difference, consider conditioning on two events. In the pairwise independent case, although any one event is independent of each of the other two individually, it is not independent of the intersection of the other two:

:<math>\mathrm{P}(A|BC) = \frac{\frac{4}{40}}{\frac{4}{40} + \frac{1}{40}} = \tfrac{4}{5} \ne \mathrm{P}(A)</math>
:<math>\mathrm{P}(B|AC) = \frac{\frac{4}{40}}{\frac{4}{40} + \frac{1}{40}} = \tfrac{4}{5} \ne \mathrm{P}(B)</math>
:<math>\mathrm{P}(C|AB) = \frac{\frac{4}{40}}{\frac{4}{40} + \frac{6}{40}} = \tfrac{2}{5} \ne \mathrm{P}(C)</math>

In the mutually independent case, however,
:<math>\mathrm{P}(A|BC) = \frac{\frac{1}{16}}{\frac{1}{16} + \frac{1}{16}} = \tfrac{1}{2} = \mathrm{P}(A)</math>
:<math>\mathrm{P}(B|AC) = \frac{\frac{1}{16}}{\frac{1}{16} + \frac{1}{16}} = \tfrac{1}{2} = \mathrm{P}(B)</math>
:<math>\mathrm{P}(C|AB) = \frac{\frac{1}{16}}{\frac{1}{16} + \frac{3}{16}} = \tfrac{1}{4} = \mathrm{P}(C)</math>

===Triple-independence but no pairwise-independence===
It is possible to create a three-event example in which

:<math>\mathrm{P}(A \cap B \cap C) = \mathrm{P}(A)\mathrm{P}(B)\mathrm{P}(C),</math>

and yet no two of the three events are pairwise independent (and hence the set of events are not mutually independent).<ref>George, Glyn, "Testing for the independence of three events," ''Mathematical Gazette'' 88, November 2004, 568. [http://www.engr.mun.ca/~ggeorge/MathGaz04.pdf PDF]</ref> This example shows that mutual independence involves requirements on the products of probabilities of all combinations of events, not just the single events as in this example.