(2nd International Conference on Teaching Statistics, Victoria (Canada))

1. For Random events we consider the notions of a mutual independence and pairwise independence.

2. Suppose A, B, C are random events satisfying the relation:

P(ABC)=P(A)P(B)P(C).

SOLUTION 1. (BERNSTEIN, 1928)

SUPPOSE A BOX CONTAINS 4 TICKETS LABELLED BY

LET US CHOOSE ONE TICKET AT RANDOM, AND CONSIDER THE RANDOM EVENTS

A₁={1 OCCURS AT THE FIRST PLACE}

A₂={1 OCCURS AT THE SECOND PLACE}

A₃={1 OCCURS AT THE THIRD PLACE}

P(A₁)=1/2 P(A₂)=1/2 P(A₃)=1/2

A₁A₂={112} A₁A₃={121} A₂A₃={211}

P(A₁A₂)=P(A₁A₃)=P(A₂A₃)=1/4.

So we conclude that the three events A₁, A₂, A₃ are pairwise independent.

However

CONCLUSION: Pairwise independence of a given set of random events does not imply that these events are mutually independent.

Toss two different standard dice, white and black.

The sample space S of the outcomes consists of all oredred pairs

ij, i,j=1, ..., 6

S={(1,1), (1,2), ..., (6,6)}

Each point in S has probability 1/36.

A₁={first die=1,2 or 3}

A₂={first die=3,4 or 5}

A₃={sum of faces is 9}

A₁A₂={31,32,33,34,35,36}

A₁A₃={36}

A₂A₃={36,45,54}

A₁A₂A₃={36}

P(A₁)=1/2 P(A₂)=1/2 P(A₃)=1/9

P(A₁A₂A₃)=1/36=(1/2)(1/2)(1/9)=P(A₁)P(A₂)P(A₃)

CONCLUSION: Mutual independence of a given set of random events does not imply that these events are pairwise independent.