Exercise 16 Page 867

We want to find the expected value of drawing one marble and decide whether we should play this game for $\$5.$ Let's first recall the definition of expected value.

Expected Value

If $A$ is an action that includes outcomes $A_1,$ $A_2,$ $A_3,\dots$ and $\text{Value}(A_n)$ is a quantitative value associated with each outcome, the expected value of $A$ is given by $\text{Value}(A)=P(A_1)\cdot \text{Value}(A_1)+P(A_2)\cdot \text{Value}(A_2)+\dots$

In other words, it is what we get, when we add up all products of the possible amount of earned money and the corresponding probability. Therefore, we begin by calculating the probability of drawing red, yellow, and blue marble. Let's use the theoretical probability. A bag contains $10$ marbles, so in each case the number of possible outcomes is equal to $10.$ We also know there are $4$ red, $5$ yellow, and $1$ blue marbles, therefore we have enough information to calculate all probabilities. Now, let's organize the obtained information in the table.

	Distribution of Prizes
Money Prize, $\text{Value}(A_i)$	$\$10$ (blue)	$\$5$ (red)	$\$3$ (yellow)
Probability, $P(A_i)$	$\frac{1}{10}=0.1$	$\frac{4}{10}=0.4$	$\frac{5}{10}=0.5$

Now, we can calculate all the products of the possible prizes and their probabilities.

	Distribution of Prizes
Money Prize, $\text{Value}(A_i)$	$\$10$ (blue)	$\$5$ (red)	$\$3$ (yellow)
Probability, $P(A_i)$	$\frac{1}{10}=0.1$	$\frac{4}{10}=0.4$	$\frac{5}{10}=0.5$
$P(A_i)\cdot \text{Value}(A_i)$	$0.1\cdot \$10=\$1$	$0.4\cdot \$5=\$2$	$0.5\cdot \$3=\$1.5$

Finally, we can calculate the expected value by summing all values from the the last row. Let's do it! The expected value is equal to $\$4.5,$ so we conclude that on average we are expected to win $\$4.5$ in this game. Therefore, we should not play this game for $\$5,$ because the expected value is less than $\$5.$