Chapter 2.6, Problem 8E

To determine

Show that events $E_1,E_2,E_3$ are pairwise independent.

Explanation of Solution

Calculation:

Probability:

The formula for probability of event E is,

$P\left(E\right)=\frac{N\left(E\right)}{N}$

In the formula, $N\left(E\right)$ is the number of outcomes for event E and N is the total number of outcomes.

Pairwise independence:

Let $E_1$, $E_2$, $E_3$ be three events, then events are said to be pairwise independent if,

$\begin{array}{l}P\left(E_1\cap E_2\right)=P\left(E_1\right)P\left(E_2\right)\\ P\left(E_1\cap E_3\right)=P\left(E_1\right)P\left(E_3\right)\\ P\left(E_2\cap E_3\right)=P\left(E_2\right)P\left(E_3\right)\end{array}$

Also, pairwise independence does not imply $P\left(E_1\cap E_2\cap E_3\right)=P\left(E_1\right)P\left(E_2\right)P\left(E_3\right)$.

The event $E_1$ is defined as sum of the two outcomes is 7 and event $E_2$ is defined as outcome of the first roll is 3 and event $E_3$ is defined as outcome of the second roll is 4.

A die has 6 outcomes $\left\{1,2,3,4,5,6\right\}$. When a die is rolled twice the outcomes are,

$S=\left\{\begin{array}{l}\left(1,1\right),\left(1,2\right),\left(1,3\right),\left(1,4\right),\left(1,5\right),\left(1,6\right)\\ \left(2,1\right),\left(2,2\right),\left(2,3\right),\left(2,4\right),\left(2,5\right),\left(2,6\right)\\ \left(3,1\right),\left(3,2\right),\left(3,3\right),\left(3,4\right),\left(3,5\right),\left(3,6\right)\\ \left(4,1\right),\left(4,2\right),\left(4,3\right),\left(4,4\right),\left(4,5\right),\left(4,6\right)\\ \left(5,1\right),\left(5,2\right),\left(5,3\right),\left(5,4\right),\left(5,5\right),\left(5,6\right)\\ \left(6,1\right),\left(6,2\right),\left(6,3\right),\left(6,4\right),\left(6,5\right),\left(6,6\right)\end{array}\right\}$

The total possible outcomes are, $N=36$.

The favorable outcomes for event $E_1$ are $\left\{\left(1,6\right),\left(2,5\right),\left(3,4\right),\left(4,3\right),\left(5,2\right),\left(6,1\right)\right\}$. The probability is,

$\begin{array}{c}P\left(E_1\right)=\frac{6}{36}\\ =0.1667\end{array}$

The favorable outcomes for event $E_2$ are $\left\{\left(3,1\right),\left(3,2\right),\left(3,3\right),\left(3,4\right),\left(3,5\right),\left(3,6\right)\right\}$. The probability is,

$\begin{array}{c}P\left(E_2\right)=\frac{6}{36}\\ =0.1667\end{array}$

The favorable outcomes for event $E_3$ are $\left\{\left(1,4\right),\left(2,4\right),\left(3,4\right),\left(4,4\right),\left(5,4\right),\left(6,4\right)\right\}$. The probability is,

$\begin{array}{c}P\left(E_3\right)=\frac{6}{36}\\ =0.1667\end{array}$

The favorable outcome for event $E_1\cap E_2$ is $\left(3,4\right)$. The probability is,

$\begin{array}{c}P\left(E_1\cap E_2\right)=\frac{1}{36}\\ =0.0278\end{array}$

The favorable outcome for event $E_1\cap E_3$ is $\left(3,4\right)$. The probability is,

$\begin{array}{c}P\left(E_1\cap E_3\right)=\frac{1}{36}\\ =0.0278\end{array}$

The favorable outcome for event $E_2\cap E_3$ is $\left(3,4\right)$. The probability is,

$\begin{array}{c}P\left(E_2\cap E_3\right)=\frac{1}{36}\\ =0.0278\end{array}$

Independence of events $E_1$ and $E_2$:

$\begin{array}{c}P\left(E_1\right)P\left(E_2\right)=0.1667\times 0.1667\\ =0.0278\\ =P\left(E_1\cap E_2\right)\end{array}$

Independence of events $E_1$ and $E_3$:

$\begin{array}{c}P\left(E_1\right)P\left(E_3\right)=0.1667\times 0.1667\\ =0.0278\\ =P\left(E_1\cap E_3\right)\end{array}$

Independence of events $E_2$ and $E_3$:

$\begin{array}{c}P\left(E_2\right)P\left(E_3\right)=0.1667\times 0.1667\\ =0.0278\\ =P\left(E_2\cap E_3\right)\end{array}$

It can be observed that, the events $E_1,E_2,E_3$ are pairwise independent.

The favorable outcome for $E_1\cap E_2\cap E_3$ is $\left(3,4\right)$. The probability is,

$\begin{array}{c}P\left(E_1\cap E_2\cap E_3\right)=\frac{1}{36}\\ =0.0278\end{array}$

Independence of events $E_1$, $E_2$ and $E_3$:

$\begin{array}{c}P\left(E_1\right)P\left(E_2\right)P\left(E_3\right)=0.1667\times 0.1667\times 0.1667\\ =0.0046\\ \ne P\left(E_1\cap E_2\cap E_3\right)\end{array}$

Hence, the pairwise independence does not imply $P\left(E_1\cap E_2\cap E_3\right)=P\left(E_1\right)P\left(E_2\right)P\left(E_3\right)$.