Question

10.1 notetaking with vocabulary (continued)
extra practice
in exercises 1 and 2, find the number of possible outcomes in the sample space.
then list the possible outcomes.

1. a stack of cards contains the thirteen clubs from a standard deck of cards. you pick a card

from the stack and flip two coins.

1. you spin a spinner with the numbers 1–5 on it and roll a die.

1. when two tiles with numbers between 1 and 10 are chosen from two different bags,

there are 100 possible outcomes. find the probability that (a) the sum of the two
numbers is not 10 and (b) the product of the numbers is greater than 10.

1. at a school dance, the parents sell pizza slices. the table shows the number of pizza

slices that are available. a student chooses a slice at random. what is the probability
that the student chooses a thin crust slice with pepperoni?

	pepperoni	plain cheese
thick crust	8	12

Explanation:

Problem 1:

Step1: Determine outcomes for card

There are 13 club cards, so 13 outcomes for picking a card.

Step2: Determine outcomes for coins

Flipping two coins: each coin has 2 outcomes, so total for two coins is \(2\times2 = 4\) outcomes (HH, HT, TH, TT).

Step3: Calculate total outcomes

Using the multiplication principle, total outcomes = \(13\times4 = 52\).

Step4: List possible outcomes

For each club card (Ace, 2, 3, ..., 10, Jack, Queen, King of clubs), combine with each coin flip outcome. For example: (Ace of clubs, HH), (Ace of clubs, HT), (Ace of clubs, TH), (Ace of clubs, TT), (2 of clubs, HH), etc.

Step1: Determine outcomes for spinner

The spinner has numbers 1 - 5, so 5 outcomes.

Step2: Determine outcomes for die

A die has 6 faces, so 6 outcomes.

Step3: Calculate total outcomes

Using the multiplication principle, total outcomes = \(5\times6 = 30\).

Step4: List possible outcomes

For each number on the spinner (1, 2, 3, 4, 5), combine with each number on the die (1, 2, 3, 4, 5, 6). For example: (1, 1), (1, 2), ..., (1, 6), (2, 1), ..., (5, 6).

Step1: Find total number of outcomes

We know total outcomes \(n = 100\) (since we choose from two bags with numbers 1 - 10, \(10\times10=100\)).

Step2: Find number of outcomes where sum is 10

Let the numbers be \(x\) (from first bag) and \(y\) (from second bag), \(x + y=10\), \(x,y\in\{1,2,\cdots,10\}\). The pairs \((x,y)\) are: (1,9), (2,8), (3,7), (4,6), (5,5), (6,4), (7,3), (8,2), (9,1). So there are 9 outcomes where the sum is 10.

Step3: Find number of outcomes where sum is not 10

Using the formula \(n(\text{not }A)=n - n(A)\), where \(A\) is the event that sum is 10. So \(n(\text{not }A)=100 - 9 = 91\).

Step4: Calculate probability

Probability \(P(\text{sum not }10)=\frac{n(\text{not }A)}{n}=\frac{91}{100}=0.91\).

Answer:

Number of possible outcomes: 52. Possible outcomes: Each club card (13 total) paired with each of the 4 coin - flip results (HH, HT, TH, TT), e.g., (Ace of clubs, HH), (Ace of clubs, HT), ..., (King of clubs, TT).

Problem 2: