Exercise 13 Page 808

Practice makes perfect

a We can see that both the die and the spinner have numbers from

1

6

as outcomes. Therefore, the possible sums obtained from rolling a die and spinning the spinner go from

2

12 .

Let's look at the spinner!

Looking at the figure, we can see that sections $1$ and $4$ are twice as big as the sections of the other numbers. We can see this as the regions of $1$ and $4$ having two sections in the spinner. Let's write the probabilities of the spinner landing on each section.

Outcome	Probability
$1$	$\frac{2}{8} = \frac{1}{4}$
$2$	$\frac{1}{8}$
$3$	$\frac{1}{8}$
$4$	$\frac{2}{8} = \frac{1}{4}$
$5$	$\frac{1}{8}$
$6$	$\frac{1}{8}$

We know that all the outcomes of a die have a probability of $\frac{1}{6} .$ Let's write the ways in which we can get every possible sum. We will write the numbers $1$ and $4$ of the spinner in $red$ and the other numbers in $blue$ to distinguish them from the numbers of the die.

Sum	Die $+$ Spinner
$2$	$1 + 1$
$3$	$1 + 2, 2 + 1$
$4$	$1 + 3, 2 + 2, 3 + 1$
$5$	$1 + 4, 2 + 3, 3 + 2, 4 + 1$
$6$	$1 + 5, 2 + 4, 3 + 3, 4 + 2, 5 + 1$
$7$	$1 + 6, 2 + 5, 3 + 4, 4 + 3, 5 + 2, 6 + 1$
$8$	$2 + 6, 3 + 5, 4 + 4, 5 + 3, 6 + 2$
$9$	$3 + 6, 4 + 5, 5 + 4, 6 + 3$
$10$	$4 + 6, 5 + 5, 6 + 4$
$11$	$5 + 6, 6 + 5$
$12$	$6 + 6$

The events of rolling a die and spinning the spinner are independent. Let's find the probability of rolling a number and spinning a red number.

\frac{1}{6} \cdot \frac{2}{8} = \frac{2}{4 8}

Similarly, let's find the probability of rolling a number and spinning a blue number.

\frac{1}{6} \cdot \frac{1}{8} = \frac{1}{4 8}

Let's substitute these values for every combination to find the probability.

Sum	Probability
$2$	$\frac{2}{4 8}$
$3$	$\frac{1}{4 8} + \frac{2}{4 8} = \frac{3}{4 8}$
$4$	$\frac{1}{4 8} + \frac{1}{4 8} + \frac{2}{4 8} = \frac{4}{4 8}$
$5$	$\frac{2}{4 8} + \frac{1}{4 8} + \frac{1}{4 8} + \frac{2}{4 8} = \frac{6}{4 8}$
$6$	$\frac{1}{4 8} + \frac{2}{4 8} + \frac{1}{4 8} + \frac{1}{4 8} + \frac{2}{4 8} = \frac{7}{4 8}$
$7$	$\frac{1}{4 8} + \frac{1}{4 8} + \frac{2}{4 8} + \frac{1}{4 8} + \frac{1}{4 8} + \frac{2}{4 8} = \frac{8}{4 8}$
$8$	$\frac{1}{4 8} + \frac{1}{4 8} + \frac{2}{4 8} + \frac{1}{4 8} + \frac{1}{4 8} = \frac{6}{4 8}$
$9$	$\frac{1}{4 8} + \frac{1}{4 8} + \frac{2}{4 8} + \frac{1}{4 8} = \frac{5}{4 8}$
$10$	$\frac{1}{4 8} + \frac{1}{4 8} + \frac{2}{4 8} = \frac{4}{4 8}$
$11$	$\frac{1}{4 8} + \frac{1}{4 8} = \frac{2}{4 8}$
$12$	$\frac{1}{4 8}$

We can see that a sum of $12$ is the least probable. Therefore, this event has a frequency of $1 .$ We can see that the numerators of the probability of each event indicates the frequency. Let's make the relative-frequency table.

$X =$ Sum	Frequency	Probability
$2$	$2$	$\frac{1}{2 4}$
$3$	$3$	$\frac{1}{1 6}$
$4$	$4$	$\frac{1}{1 2}$
$5$	$6$	$\frac{1}{8}$
$6$	$7$	$\frac{7}{4 8}$
$7$	$8$	$\frac{1}{6}$
$8$	$6$	$\frac{1}{8}$
$9$	$5$	$\frac{5}{4 8}$
$10$	$4$	$\frac{1}{1 2}$
$11$	$2$	$\frac{1}{2 4}$
$12$	$1$	$\frac{1}{4 8}$

b Let's make a bar graph with the probabilities we found in Part A. The bar is associated with one of the sums. The height of each bar represents the probability of the event.

c Let's recall the formula for the expected value of a probability distribution.

E (X) = [X_{1} \cdot P (X_{1})] + [X_{2} \cdot P (X_{2})] + \dots + [X_{n} \cdot P (X_{n})]

In this formula,

X_{n}

is an outcome and

P (X_{n})

is the probability of outcome

X_{n} .

Let's substitute the values from the table of Part A into the formula to find the expected value of one spin.

E (X) = [X_{1} \cdot P (X_{1})] + \dots + [X_{n} \cdot P (X_{n})]

SubstituteValues

Substitute values

E (X) = [2 \cdot \frac{1}{2 4}] + [3 \cdot \frac{1}{1 6}] + [4 \cdot \frac{1}{1 2}] + [5 \cdot \frac{1}{8}] + [6 \cdot \frac{7}{4 8}] + [7 \cdot \frac{1}{6}] + [8 \cdot \frac{1}{8}] + [9 \cdot \frac{5}{4 8}] + [10 \cdot \frac{1}{1 2}] + [11 \cdot \frac{1}{2 4}] + [12 \cdot \frac{1}{4 8}]

Multiply

E (X) = \frac{1}{1 2} + \frac{3}{1 6} + \frac{1}{3} + \frac{5}{8} + \frac{7}{8} + \frac{7}{6} + 1 + \frac{1 5}{1 6} + \frac{5}{6} + \frac{1 1}{2 4} + \frac{1}{4}

AddTerms

Add terms

E (X) = 6.75

d From Part A we already know each outcome and its theoretical probability. For the probability model we will use a random number generator. We will assign each integer from

1

48

to represent the probability of each outcome. Let's write an example.

Outcome	Range
$2$	$1 - 2$

Each number from $1$ to $48$ has a $\frac{1}{4 8}$ chance of being generated. Therefore, since the outcome $2$ has a probability of $\frac{2}{4 8},$ we assign two numbers from the interval to this outcome. Let's write the complete table!

Outcome	Range
$2$	$1$ - $2$
$3$	$3$ - $5$
$4$	$6$ - $9$
$5$	$10$ - $15$
$6$	$16$ - $22$
$7$	$23$ - $30$
$8$	$31$ - $36$
$9$	$37$ - $41$
$10$	$42$ - $45$
$11$	$46$ - $47$
$12$	$48$

The trial will represent selecting a number at random. To conduct the simulation, we will use a graphing calculator. To do so, we need to push the $MATH$ button. Then, we will scroll to the right to the PRB menu and choose the fifth option, randInt(.

The function randInt $(a, b, c)$ generates $c$ random integers from $a$ to $b,$ inclusive. Therefore, we will evaluate randInt $(1, 48, 50) .$ Then, we will use the left and right arrow buttons to see the results.

Now we will write the frequency of our results in the table.

Outcome	Range	Tally
$2$	$1$ - $2$	$4$
$3$	$3$ - $5$	$2$
$4$	$6$ - $9$	$4$
$5$	$10$ - $15$	$10$
$6$	$16$ - $22$	$5$
$7$	$23$ - $30$	$10$
$8$	$31$ - $36$	$5$
$9$	$37$ - $41$	$2$
$10$	$42$ - $45$	$4$
$11$	$46$ - $47$	$4$
$12$	$48$	$0$

e Let's make a graph with the results of the simulation from Part D.

This graph illustrates the experimental probability, while the graph from Part B illustrates the theoretical probability. We can see that even if the graphs differ from each other, the graph of the experimental probability shares some similarities, like having most of the trials on the left side.

Subchapter links

Check Your Understanding p.806

Practice and Problem Solving p.806-808

H.O.T. Problems p.808

Standardized Test Practice p.809

Spiral Review p.809

Exercise 13 Page 808

Hints

Check the answer

Practice exercises

Asses your skill