Course 3 10-1 Probability Students will learn to find the probability of an event by using the definition of probability.

Course 3 10-1 Probability Normal x x

Course 3 10-1 Probability

10-1 Probability Standard Deviation Course 3 10-1 Probability Standard Deviation The standard deviation of n data values is the square root of the mean of the squared deviations from the distribution's mean. If x1, x2, ..., xn are data values from a sample, then the standard deviation s is given by:

Course 3 10-1 Probability

Combination C-Combination Combination

10-1 Probability Picking from 52 cards of a bridge deck Course 3 10-1 Probability An experiment is an activity in which results are observed. Each observation is called a trial, and each result is called an outcome. The sample space is the set of all possible outcomes of an experiment. Experiment Sample Space flipping a coin heads, tails rolling a number cube 1, 2, 3, 4, 5, 6 guessing the number of whole numbers marbles in a jar Picking from 52 cards of a bridge deck

Course 3 10-1 Probability An event is any set of one or more outcomes. The probability of an event, written P(event), is a number from 0 (or 0%) to 1 (or 100%) that tells you how likely the event is to happen. A probability of 0 means the event is impossible, or can never happen. A probability of 1 means the event is certain, or has to happen. The probabilities of all the outcomes in the sample space add up to 1.

Probability Concepts 10-1 Probability Course 3 10-1 Probability Probability Concepts Independent Event: Occurrence of one does not influence the probability of occurrence of the other Dependent: Occurrence of one affects the probability of the other Independent Events E1 = heads on one flip of fair coin E2 = heads on second flip of same coin Result of second flip does not depend on the result of the first flip. Dependent Events E1 = rain forecasted on the news E2 = take umbrella to work Probability of the second event is affected by the occurrence of the first event

Course 3 10-1 Probability

Visualizing Events 10-1 Probability Contingency Tables Tree Diagrams Course 3 10-1 Probability Visualizing Events Contingency Tables Tree Diagrams Ace Not Ace Total Black 2 24 26 Red 2 24 26 Total 4 48 52 Sample Space 2 24 Ace Sample Space Black Card Not an Ace Full Deck of 52 Cards Ace Red Card Not an Ace

Example 1: Finding Probabilities of Outcomes in a Sample Space Course 3 10-1 Probability Example 1: Finding Probabilities of Outcomes in a Sample Space Give the probability for each outcome. The basketball team has a 70% chance of winning. The probability of winning is P(win) = 70% = 0.7. The probabilities must add-up to 1 (100), so the probability of not winning is q(lose) = 1 – 0.7 = 0.3, or 30%.

10-1 Probability Example 1: You try it… Course 3 10-1 Probability Example 1: You try it… Give the probability for each outcome. The polo team has a 50% chance of winning. The probability of winning is P(win) = 50% = 0.5. The probabilities must addup to 1, so the probability of not winning is q(lose) = 1 – 0.5 = 0.5, or 50%.

Example 2: Finding Probabilities of Outcomes in a Sample Space Course 3 10-1 Probability Example 2: Finding Probabilities of Outcomes in a Sample Space Give the probability for each outcome. 3 2 3 8 8 8 3 8 2 + = 1 

10-1 Probability 1 2 3 4 5 6  Example 2: You Try it… Outcome Course 3 10-1 Probability Example 2: You Try it… Give the probability for each outcome. Rolling a number cube. Outcome 1 2 3 4 5 6 Probability 1 6 + = 1 

Course 3 10-8 Counting Principles Example 3: Using the Fundamental Counting Principle License plates are being produced that have a single letter followed by three digits. All license plates are equally likely. Find the number of possible license plates. Use the Fundamental Counting Principal. letter first digit second digit third digit 26 choices 10 choices 10 choices 10 choices 26 • 10 • 10 • 10 = 26,000 The number of possible 1-letter, 3-digit license plates is 26,000.

Course 3 10-1 Probability To find the probability of an event, add the probabilities of all the outcomes included in the event.

Additional Example 4: Using the Fundamental Counting Principal Course 3 10-8 Counting Principles Additional Example 4: Using the Fundamental Counting Principal Find the probability that a license plate has the letter Q. 1 26  1 • 10 • 10 • 10 26,000 = P(Q ) = 0.038

Additional Example 5: Using the Fundamental Counting Principle Course 3 10-8 Counting Principles Additional Example 5: Using the Fundamental Counting Principle Find the probability that a license plate does not contain a 3. First use the Fundamental Counting Principle to find the number of license plates that do not contain a 3. 26 • 9 • 9 • 9 = 18,954 possible license plates without a 3 There are 9 choices for any digit except 3. P(no 3) = = 0.729 26,000 18,954

Example 3: You Try it Continued… Course 3 10-8 Counting Principles Example 3: You Try it Continued… Social Security numbers contain 9 digits. All social security numbers are equally likely. Find the number of possible Social Security numbers. Use the Fundamental Counting Principle. Digit 1 2 3 4 5 6 7 8 9 Choices 10 10 • 10 • 10 • 10 • 10 • 10 • 10 • 10 • 10 = 1,000,000,000 The number of Social Security numbers is 1,000,000,000.

Example 3B: You Try it Continued… Course 3 10-8 Counting Principles Example 3B: You Try it Continued… Find the probability that the Social Security number contains a 7. P(7 _ _ _ _ _ _ _ _) = 1 • 10 • 10 • 10 • 10 • 10 • 10 • 10 • 10 1,000,000,000 = = 0.1 10 1

Example 3C: You Try it Continued… Course 3 10-8 Counting Principles Example 3C: You Try it Continued… Find the probability that a Social Security number does not contain a 7. First use the Fundamental Counting Principle to find the number of Social Security numbers that do not contain a 7. P(no 7 _ _ _ _ _ _ _ _) = 9 • 9 • 9 • 9 • 9 • 9 • 9 • 9 • 9 1,000,000,000 P(no 7) = ≈ 0.4 1,000,000,000 387,420,489

Course 3 10-8 Counting Principles The Fundamental Counting Principle tells you only the number of outcomes in some experiments, not what the outcomes are. A tree diagram is a way to show all of the possible outcomes.

Example 6: Using a Tree Diagram Course 3 10-8 Counting Principles Example 6: Using a Tree Diagram You have a photo that you want to mat and frame. You can choose from a blue, purple, red, or green mat and a metal or wood frame. Describe all of the ways you could frame this photo with one mat and one frame. You can find all of the possible outcomes by making a tree diagram. There should be 4 • 2 = 8 different ways to frame the photo.

Example 6: Using a Tree Diagram Continued… Course 3 10-8 Counting Principles Example 6: Using a Tree Diagram Continued… Each “branch” of the tree diagram represents a different way to frame the photo. The ways shown in the branches could be written as (blue, metal), (blue, wood), (purple, metal), (purple, wood), (red, metal), (red, wood), (green, metal), and (green, wood).

Counting Principles 10-8 Example 6: You Try it Course 3 10-8 Counting Principles Example 6: You Try it A baker can make yellow or white cakes with a choice of chocolate, strawberry, or vanilla icing. Describe all of the possible combinations of cakes. You can find all of the possible outcomes by making a tree diagram. There should be 2 • 3 = 6 different cakes available.

Example 6: You Try it Continued… Course 3 10-8 Counting Principles Example 6: You Try it Continued… yellow cake The different cake possibilities are (yellow, chocolate), (yellow, strawberry), (yellow, vanilla), (white, chocolate), (white, strawberry), and (white, vanilla). vanilla icing chocolate icing strawberry icing white cake vanilla icing chocolate icing strawberry icing

Additional Example 3: Using the Addition Counting Principle Course 3 10-8 Counting Principles Additional Example 3: Using the Addition Counting Principle The table shows the items available at a farm stand. How many items can you choose from the farm stand? Apples Pears Squash Macintosh Bosc Acorn Red Delicious Yellow Bartlett Hubbard Gold Delicious Red Bartlett None of the lists contains identical items, so use the Addition Counting Principle. Total Choices = Apples + Pears + Squash

Additional Example 3 Continued Course 3 10-8 Counting Principles Additional Example 3 Continued T = 3 + 3 + 2 = 8 There are 8 items to choose from.

Independent and Dependent Events Course 3 10-5 Independent and Dependent Events A compound event is made up of one or more separate events. To find the probability of a compound event, you need to know if the events are independent or dependent. Events are independent events if the occurrence of one event does not affect the probability of the other. Events are dependent events if the occurrence of one does affect the probability of the other.

Additional Example 1: Classifying Events as Independent or Dependent Course 3 10-5 Independent and Dependent Events Additional Example 1: Classifying Events as Independent or Dependent Determine if the events are dependent or independent. A. getting tails on a coin toss and rolling a 6 on a number cube B. getting 2 red gumballs out of a gumball machine Tossing a coin does not affect rolling a number cube, so the two events are independent. After getting one red gumball out of a gumball machine, the chances for getting the second red gumball have changed, so the two events are dependent.

Independent and Dependent Events Course 3 10-5 Independent and Dependent Events Check It Out: Example 1 Determine if the events are dependent or independent. A. rolling a 6 two times in a row with the same number cube B. a computer randomly generating two of the same numbers in a row The first roll of the number cube does not affect the second roll, so the events are independent. The first randomly generated number does not affect the second randomly generated number, so the two events are independent.

Independent and Dependent Events Course 3 10-5 Independent and Dependent Events

Additional Example 2A: Finding the Probability of Independent Events Course 3 10-5 Independent and Dependent Events Additional Example 2A: Finding the Probability of Independent Events Three separate boxes each have one blue marble and one green marble. One marble is chosen from each box. What is the probability of choosing a blue marble from each box? The outcome of each choice does not affect the outcome of the other choices, so the choices are independent. In each box, P(blue) = . 12 12 · = 18 = P(blue, blue, blue) = 0.125 Multiply.

Additional Example 2C: Finding the Probability of Independent Events Course 3 10-5 Independent and Dependent Events Additional Example 2C: Finding the Probability of Independent Events What is the probability of choosing at least one blue marble? Think: P(at least one blue) + P(not blue, not blue, not blue) = 1. In each box, P(not blue) = . 1 2 P(not blue, not blue, not blue) = 12 · = 18 = 0.125 Multiply. Subtract from 1 to find the probability of choosing at least one blue marble. 1 – 0.125 = 0.875

Independent and Dependent Events Course 3 10-5 Independent and Dependent Events Check It Out: Example 2A Two boxes each contain 4 marbles: red, blue, green, and black. One marble is chosen from each box. What is the probability of choosing a blue marble from each box? The outcome of each choice does not affect the outcome of the other choices, so the choices are independent. In each box, P(blue) = . 14 14 · = 1 16 = P(blue, blue) = 0.0625 Multiply.

Independent and Dependent Events Course 3 10-5 Independent and Dependent Events Check It Out: Example 2B Two boxes each contain 4 marbles: red, blue, green, and black. One marble is chosen from each box. What is the probability of choosing a blue marble and then a red marble? In each box, P(blue) = . 14 In each box, P(red) = . 14 14 · = 1 16 = P(blue, red) = 0.0625 Multiply.

Independent and Dependent Events Course 3 10-5 Independent and Dependent Events Check It Out: Example 2C Two boxes each contain 4 marbles: red, blue, green, and black. One marble is chosen from each box. What is the probability of choosing at least one blue marble? Think: P(at least one blue) + P(not blue, not blue) = 1. In each box, P(blue) = . 14 34 · = 9 16 = P(not blue, not blue) = 0.5625 Multiply. Subtract from 1 to find the probability of choosing at least one blue marble. 1 – 0.5625 = 0.4375

Independent and Dependent Events Course 3 10-5 Independent and Dependent Events To calculate the probability of two dependent events occurring, do the following: 1. Calculate the probability of the first event. 2. Calculate the probability that the second event would occur if the first event had already occurred. 3. Multiply the probabilities.

Additional Example 3A: Find the Probability of Dependent Events Course 3 10-5 Independent and Dependent Events Additional Example 3A: Find the Probability of Dependent Events The letters in the word dependent are placed in a box. If two letters are chosen at random, what is the probability that they will both be consonants? Because the first letter is not replaced, the sample space is different for the second letter, so the events are dependent. Find the probability that the first letter chosen is a consonant. 69 = 23 P(first consonant) =

Additional Example 3A Continued Course 3 10-5 Independent and Dependent Events Additional Example 3A Continued If the first letter chosen was a consonant, now there would be 5 consonants and a total of 8 letters left in the box. Find the probability that the second letter chosen is a consonant. 58 P(second consonant) = 58 23 · = 5 12 Multiply. The probability of choosing two letters that are both consonants is . 5 12

Additional Example 3B: Find the Probability of Dependent Events Course 3 10-5 Independent and Dependent Events Additional Example 3B: Find the Probability of Dependent Events If two letters are chosen at random, what is the probability that they will both be consonants or both be vowels? There are two possibilities: 2 consonants or 2 vowels. The probability of 2 consonants was calculated in Example 3A. Now find the probability of getting 2 vowels. Find the probability that the first letter chosen is a vowel. 39 = 13 P(first vowel) = If the first letter chosen was a vowel, there are now only 2 vowels and 8 total letters left in the box.

Additional Example 3B Continued Course 3 10-5 Independent and Dependent Events Additional Example 3B Continued Find the probability that the second letter chosen is a vowel. 28 = 14 P(second vowel) = 14 13 · = 1 12 Multiply. The events of both consonants and both vowels are mutually exclusive, so you can add their probabilities. 5 12 1 12 + = 6 12 = 12 P(consonant) + P(vowel) The probability of getting two letters that are either both consonants or both vowels is . 12

Independent and Dependent Events Course 3 10-5 Independent and Dependent Events Two mutually exclusive events cannot both happen at the same time. Remember!

Independent and Dependent Events Course 3 10-5 Independent and Dependent Events Check It Out: Example 3A The letters in the phrase I Love Math are placed in a box. If two letters are chosen at random, what is the probability that they will both be consonants? Because the first letter is not replaced, the sample space is different for the second letter, so the events are dependant. Find the probability that the first letter chosen is a consonant. 59 P(first consonant) =

Check It Out: Example 3A Continued Course 3 10-5 Independent and Dependent Events Check It Out: Example 3A Continued If the first letter chosen was a consonant, now there would be 4 consonants and a total of 8 letters left in the box. Find the probability that the second letter chosen is a consonant. 48 = 12 P(second consonant) = 12 59 · = 5 18 Multiply. The probability of choosing two letters that are both consonants is . 5 18

Independent and Dependent Events Course 3 10-5 Independent and Dependent Events Check It Out: Example 3B If two letters are chosen at random, what is the probability that they will both be consonants or both be vowels? There are two possibilities: 2 consonants or 2 vowels. The probability of 2 consonants was calculated in Try This 3A. Now find the probability of getting 2 vowels. Find the probability that the first letter chosen is a vowel. 49 P(first vowel) = If the first letter chosen was a vowel, there are now only 3 vowels and 8 total letters left in the box.

Check It Out: Example 3B Continued Course 3 10-5 Independent and Dependent Events Check It Out: Example 3B Continued 38 Find the probability that the second letter chosen is a vowel. P(second vowel) = 38 49 · = 12 72 16 = Multiply. The events of both consonants and both vowels are mutually exclusive, so you can add their probabilities. 5 18 1 6 + = 8 18 = 49 P(consonant) + P(vowel) The probability of getting two letters that are either both consonants or both vowels is . 49

Independent and Dependent Events Insert Lesson Title Here Course 3 10-5 Independent and Dependent Events Insert Lesson Title Here Lesson Quiz Determine if each event is dependent or independent. 1. drawing a red ball from a bucket and then drawing a green ball without replacing the first 2. spinning a 7 on a spinner three times in a row 3. A bucket contains 5 yellow and 7 red balls. If 2 balls are selected randomly without replacement, what is the probability that they will both be yellow? dependent independent 5 33