Exercise 4 Page 865

The expected value is given by the sum of each outcome's value multiplied by its probability. We first begin by giving an example. Then we will explain the meaning of the expected value.

Example of Expected Value

Consider rolling two fair dice and finding their sum. Because there are $6$ possible outcomes on each die, we can use the Fundamental Counting Principle to find the number of possible outcomes. Note that if we add the outcomes of the dice, the sum goes from $2$ to $12.$ Additionally, $2$ only occurs when we get $1$ on each die and $12$ occurs when we get two $6$. Now, let's write the number of ways each sum can occur and its probability. Also, we will multiply the probability of each outcome by the outcome's value.

Outcome $\mathbf{X}$	Probability of $\mathbf{X},$ $\mathbf{P}(\mathbf{X})$	$X\cdot P(X)$
$2$	$\frac{1}{36}$	$2\cdot \frac{1}{36}=\frac{2}{36}$
$3$	$\frac{2}{36}$	$3\cdot \frac{2}{36}=\frac{6}{36}$
$4$	$\frac{3}{36}$	$4\cdot \frac{3}{36}=\frac{12}{36}$
$5$	$\frac{4}{36}$	$5\cdot \frac{4}{36}=\frac{20}{36}$
$6$	$\frac{5}{36}$	$6\cdot \frac{5}{36}=\frac{30}{36}$
$7$	$\frac{6}{36}$	$7\cdot \frac{6}{36}=\frac{42}{36}$
$8$	$\frac{5}{36}$	$8\cdot \frac{5}{36}=\frac{40}{36}$
$9$	$\frac{4}{36}$	$9\cdot \frac{4}{36}=\frac{36}{36}$
$10$	$\frac{3}{36}$	$10\cdot \frac{3}{36}=\frac{30}{36}$
$11$	$\frac{2}{36}$	$11\cdot \frac{2}{36}=\frac{22}{36}$
$12$	$\frac{1}{36}$	$12\cdot \frac{1}{36}=\frac{12}{36}$
Sum of Values		$\frac{252}{36}=7$

Therefore, the expected value is $7.$

What is the Expected Value?

We found that the expected value of rolling two fair dice and finding their sum is $7.$ This means that if we repeat the experiment an infinite number of times, the expected value is $7.$ Similarly, for any experiment, the expected value tells us what we expect to happen if the experiment is conducted a certain number of times. This can help us to make decisions based on data.