7.2 Pairwise independent events

Three events are said to be pairwise independent if they satisfy all the following three conditions:

Illustration 33. Two dice are thrown. Let $A$ be the event "the sum of the points is $7$", $B$ the event "the first die shows $3$" and the event $C$ be "the second die shows $4$". Comment on the nature of its dependency.

Solution: The sample space on throwing a dice is,

$$S = \{ (1,1), (1,2), (1,3), (1,4), (1,5), (1,6), (2,1), (2,2), (2,3), (2,4), (2,5), (2,6), (3,1), (3,2), (3,3), (3,4), (3,5), (3,6), (4,1), (4,2), (4,3), (4,4), (4,5), (4,6), (5,1), (5,2),(5,3),(5,4),(5,5),(5,6),(6,1),(6,2),(6,3),(6,4),(6,5),(6,6)\} $$

Total number of outcomes is $36$.

Event $A$ = The sum of the numbers is $7$

$$\{ (1,6),(2,5),(3,4),(4,3),(5,2),(6,1)\} $$

Event $B$ = The number on the first die is $3$

$$\{ (3,1),(3,2),(3,3),(3,4),(3,5),(3,6)\} $$

Event $C$ = The number on the second die is $4$

$$\{ (1,4),(2,4),(3,4),(4,4),(5,4),(6,4)\} $$

$$n(A) = 6$$

$$n(B) = 6$$

$$n(C) = 6$$

Thus,

$$\begin{equation} \begin{aligned} P(A) = {6 \over {36}} = {1 \over 6} \\ P(B) = {6 \over {36}} = {1 \over 6} \\ P(C) = {6 \over {36}} = {1 \over 6} \\\end{aligned} \end{equation} $$

$A \cap B$ = The sum is $7$ and the number on the first die is $3$ = $\{ (3,4)\} $

$B \cap C$ = The number on the first die $3$ and the number on the second die is $4$ = $\{ (3,4)\} $

$A \cap C$ = The sum is $7$ and the number on the second die is $4$ = $\{ (3,4)\} $

$$\begin{equation} \begin{aligned} P(A \cap B) = {1 \over {36}} = P(A)P(B) \\ P(B \cap C) = {1 \over {36}} = P(B)P(C) \\ P(A \cap C) = {1 \over {36}} = P(A)P(C) \\\end{aligned} \end{equation} $$

Now,

$A \cap B \cap C$ = The sum is $7$ and the number on the first die is $3$ and the number on the second die is $4$

= $\{ (3,4)\} $

$$P(A \cap B \cap C) = {1 \over {36}}$$

But,

$$\begin{equation} \begin{aligned} P(A)P(B)P(C) = {1 \over 6} \times {1 \over 6} \times {1 \over 6} = {1 \over {216}} \\ P(A)P(B)P(C) \ne P(A \cap B \cap C) \\\end{aligned} \end{equation} $$

Here,

$$\begin{equation} \begin{aligned} P(A \cap B) = P(A)P(B) \\ P(B \cap C) = P(B)P(C) \\ P(A \cap C) = P(A)P(C) \\\end{aligned} \end{equation} $$

But,

$P(A)P(B)P(C) \ne P(A \cap B \cap C)$.

Hence the above set of events are pair-wise independent events.

Illustration 34. On tossing a coin twice, the following events are taken.

$A$ : Head appears on the first toss.

$B$ : Head appears on the second toss.

$C$: Both tosses yield the same outcome

Comment on the nature of its independence.

Solution: The sample space for the following event is,

$S = \{ (H,H),(H,T),(T,H),(T,T)\} $

$A$ : Head appears on the first toss = $\{ (H,H),(H,T)\} $

$B$ : Head appears on the second toss = $\{ (H,H),(T,H)\} $

$C$: Both tosses yield the same outcome = $\{ (H,H),(T,T)\} $

$$\begin{equation} \begin{aligned} P(A) = {2 \over 4} = {1 \over 2} \\ P(B) = {2 \over 4} = {1 \over 2} \\ P(C) = {2 \over 4} = {1 \over 2} \\\end{aligned} \end{equation} $$

$$\begin{equation} \begin{aligned} A \cap B = \{ (H,H)\} \\ B \cap C = \{ (H,H)\} \\ A \cap C = \{ (H,H)\} \\\end{aligned} \end{equation} $$

$$\begin{equation} \begin{aligned} P(A \cap B) = {1 \over 4} = {1 \over 2} \times {1 \over 2} = P(A)P(B) \\ P(B \cap C) = {1 \over 4} = {1 \over 2} \times {1 \over 2} = P(B)P(C) \\ P(A \cap C) = {1 \over 4} = {1 \over 2} \times {1 \over 2} = P(A)P(C) \\\end{aligned} \end{equation} $$

$$\begin{equation} \begin{aligned} P(A)P(B)P(C) = {1 \over 2} \times {1 \over 2} \times {1 \over 2} = {1 \over 8} \\ P(A)P(B)P(C) \ne P(A \cap B \cap C). \\\end{aligned} \end{equation} $$

Here,

$$\begin{equation} \begin{aligned} P(A \cap B) = P(A)P(B) \\ P(B \cap C) = P(B)P(C) \\ P(A \cap C) = P(A)P(C) \\\end{aligned} \end{equation} $$

But,

$$P(A)P(B)P(C) \ne P(A \cap B \cap C)$$

Hence they are pairwise independent.