Answer to Question #199922 in Statistics and Probability for sumon

Question

Two coins are tossed. If A is the event “head on the first coin”, B is the event “head on the second coin” and C is the event “coins fall alike”, show that the events A, B, and C are pair wise independent but not completely independent.

Answer 1

Solution:

Given, A is the event “head on the first coin”, B is the event “head on the second coin” and C is the event “coins fall alike”

A = {HH,HT}, B = {HH,TH}, C = {TT, HH}

$A\cap B=\{HH\},B\cap C=\{HH\},A\cap C=\{HH\},A\cap B\cap C=\{HH\}$

$P(A)=P(B)=P(C)=\frac12 \\P(A\cap B)=P(B\cap C)=P(A\cap C)=P(A\cap B\cap C)=\frac14$

For A and B:

$P(A).P(B)=\frac12.\frac12=\frac14 \\P(A\cap B)=\frac14$

Since, $P(A).P(B)=P(A\cap B)$, A and B are pair-wise independent events.

For C and B:

$P(C).P(B)=\frac12.\frac12=\frac14 \\P(C\cap B)=\frac14$

Since, $P(C).P(B)=P(C\cap B)$, C and B are pair-wise independent events.

For A and B:

$P(A).P(C)=\frac12.\frac12=\frac14 \\P(A\cap C)=\frac14$

Since, $P(A).P(C)=P(A\cap C)$, A and C are pair-wise independent events.

For A,B and C:

$P(A).P(B).P(C)=\frac12.\frac12.\frac12=\frac18 \\P(A\cap B\cap C)=\frac14$

Since, $P(A).P(B).P(C)\ne P(A\cap B\cap C)$, A,B and C are not completely independent events.