CHAPTER 12: PROBABILITY AND STATISTICS

CHAPTER 12: PROBABILITY AND STATISTICS
Slide 1

Section 12-1 THE COUNTING PRINCIPLE
Thursday Definitions: Outcome-the result of a single trial. Ex. Flipping a coin has two outcomes (head or tail). Sample Space-the set of all possible outcomes. Event-consists of one or more outcomes of a trial. Independent Events-two events are independent if the occurrence (or non-occurrence) of one of the events does not affect the probability of the occurrence of the other event. Ex. The choices of letters and digits to be put on a license plate, because each letter or digit chosen does not affect the choices for the others.

Probability experiment:
12-1 Example Probability experiment: Roll a die Sample space: { } Event: { Die is even } = { } Outcome: {3}

Section 12-1 THE COUNTING PRINCIPLE
5) Fundamental Counting Principle-If there are “a” ways for one activity to occur and “b” ways for a second activity to occur, then there are (a × b) ways for both to occur. Ex. There are 6 ways to roll a die and 2 ways to flip a coin… 6 × 2 = 12 ways for both to occur. 6) Dependent events-the outcome of one event DOES affect the out come of another event.

12-1: Example: When two coins are tossed, what is the probability that both are tails?
Tree Diagram First Coin H T / \ / \ Second Coin H T H T | | | | Possible HH HT TH TT Outcomes Therefore, 4 possible outcomes. Fundamental Counting Principle: 2 possible outcomes first coin X 2 possible outcomes second coin = 4 possible outcomes P(both tails)= 1 4

12-1: Independent and Dependent Events
Independent events: if one event occurs, it does not affect the probability of the other event Drawing cards from two decks Dependent events: if one event affects the outcome of the second event, changing the probability Drawing two cards in succession from same deck without replacement

12-1: Examples Independent Events A = Being female
B = Having type O blood A = 1st child is a boy B = 2nd child is a boy Dependent Events A = taking an aspirin each day B = having a heart attack A = being a female B = being under 64” tall

12-1: Examples Determine if the following events are independent or dependent. cars are on a production line where 5 are defective and 2 cars are selected at random. A = first car is defective B = second car is defective. Two dice are rolled. A = first is a 4 and B = second is a 4 Skills Practice Workbook pg. 79 (#1-11) SG&I workbook pg. 157 (#1-8)

12-3: PROBABILITY Friday Definition: -Probability is the measure of how likely something will occur. -It is the ratio of desired outcomes to total outcomes. PROBABILITY= #desired #total -Probabilities MUST be simplified if possible.

12-3: EXAMPLE Flip a coin… -What is the probability you get heads? -What is the probability you get tails? (Remember think of all the possible outcomes)

12-3: Tree Diagrams Start 6 possible outcomes on die #1
Two six-sided dice are rolled. Describe the sample space. Start 1st roll 1 2 3 4 5 6 2nd roll 6 possible outcomes on die #1 X 6 possible outcomes on die #2 = 36outcomes

12-3: Sample Space and Probabilities
Two dice are rolled and the sum is noted. 1,1 1,2 1,3 1,4 1,5 1,6 2,1 2,2 2,3 2,4 2,5 2,6 3,1 3,2 3,3 3,4 3,5 3,6 4,1 4,2 4,3 4,4 4,5 4,6 5,1 5,2 5,3 5,4 5,5 5,6 6,1 6,2 6,3 6,4 6,5 6,6 Find the probability the sum is 4. 3/36 = 1/12 = 0.083 Find the probability the sum is 11. 2/36 = 1/18 = 0.056 Find the probability the sum is 4 or 11. 5/36 = 0.139

12-3: Probability of Success & Failure
Definitions: 1-Success-desired outcome 2-Failure-any other outcome 3-Random-when all outcomes have an equally likely chance of occurring 4-Odds-another way to measure the chance of an event occurring, it is the ratio of the number of successes to the number of failures. Ex. Success:failure

12-3: PROBABILITY VS. ODDS Ex. What is the probability of rolling a die and getting the number 5? Probability = #desired Probability=1 total # Now what are the odds? ODDS = success:failure ODDS= 1:5 (means one chance of getting a number 5 : five chances of not getting a number 5.)

12-3: Odds Odds can be written as a fraction or ratio (but must be simplified just like probabilities): Success Failures Success:Failures Note: In sports we often look at wins over losses.

12-3: Converting from Probability to ODDS EXAMPLE: Flipped a coin four times… P(two tails)= ¼ = .25 or 25% chance What are the odds? ODDS= 1 success 3 failures 1:3

12-3: PROBABILITY DISTRIBUTIONS
Monday Definitions: 1-random variable-a variable whose value is the numerical outcome of a random event. Ex. When rolling a die we can let the random variable “D” represent the number showing on the die. So, “D” can equal 1, 2, 3, 4, 5, or 6.

2-probability distribution-for a particular random variable it is a function that maps the sample space to the probabilities of the outcomes in the sample space. Ex. # on Die 1 2 3 4 5 6 Probability 1/6

3-uniform distribution-a distribution where all of the probabilities are the same. 4-relative-frequency histogram-a table of probabilities or graph that helps to visualize a probability distribution.

12-3: Probability Probability is based on observations or experiments. THERE ARE TWO TYPES OF PROBABILITY: 1- Experimental AND 2-Theoretical Definitions: An experimental probability is one that happens as the result of an experiment. # outcomes # trials ***The probability we have done so far has been “theoretical probabilities,” because there was no experiment. **Except when you do a project!

If P(E) = 0, then event E is impossible.
12-3: THEORETICAL AND EXPERIMENTAL PROBABILITY The probability of an event is a number between 0 and 1 that indicates the likelihood the event will occur. If P(E) = 0, then event E is impossible. If P(E) = 1, then event E is certain. 0  P(E)  1 Impossible Even Certain

12-3: THEORETICAL AND EXPERIMENTAL PROBABILITY
THE THEORETICAL PROBABILITY OF AN EVENT The theoretical probability of an event is often simply called the probability of the event. When all outcomes are equally likely, the theoretical probability that an event A will occur is: number of outcomes in A P (A) = 4 9 P (A) = total number of outcomes all possible outcomes outcomes in event A You can express a probability as a fraction, a decimal, or a percent. For example: , 0.5, or 50%. 1 2

12-3: EXAMPLE Theoretical
P(A) = number if ways A can occur total number of outcomes In a bag you have 3 red marbles, 2 blue marbles and 7 yellow marbles. If you select one marble at random, P(red) =

12-3: EXAMPLE Theoretical
P(A) = number if ways A can occur total number of outcomes In a bag you have 3 red marbles, 2 blue marbles and 7 yellow marbles. If you select one marble at random, P(red) = 3 / (3+2+7) = 3/12 = 1/4

12-3: Experimental examples
Prentice went fishing at a pond that contains three types of fish: blue gills, red gills and crappies. He caught 40 fish and recorded the type. The following frequency distribution shows his results. Fish Type Number of times caught Blue gill 13 Red gill 17 Crappy 10 If you catch a fish, what is the probability that it is a blue gill? A red gill? A crappy?

12-3: Subjective Probability
Subjective probability results from educated guesses, intuition and estimates. Examples… A doctor’s prediction that a patient has a 90% chance of full recovery A business analyst predicting an employee strike being 0.25

12-3: Summary Classical (Theoretical) Empirical (Statistical)
The number of outcomes in a sample space is known and each outcome is equally likely to occur. Empirical (Statistical) A.K.A. Experimental The frequency of outcomes in the sample space is estimated from experimentation. Subjective (Intuition) Probabilities result from intuition, educated guesses, and estimates. COMPLETE PROBABILITY AND ODDS WORKSHEET

12-4: Independent Events P(A∩B) = P(A) ● P(B)
Tuesday Whatever happens in one event has absolutely nothing to do with what will happen next because: The two events are unrelated OR You repeat an event with an item whose numbers will not change (ex. spinners or dice) You repeat the same activity, but you REPLACE the item that was removed. The probability of two independent events, A and B, is equal to the probability of event A times the probability of event B. P(A∩B) = P(A) ● P(B) Slide 28

12-4: Multiplication Rule for Independent Events
To get probability of both events occurring, multiply probabilities of individual events Ace from first deck and spade from second Probability of ace is 4/52 = 1/13 Probability of spade is 13/52 = 1/4 Probability of both is 1/13 x 1/4 = 1/52

12-4: INDEPENDENT EVENTS Practice: Roll a die and flip a coin: 1-P(heads and 6) = 2-P(tails and a 5) =

12-4: INDEPENDENT EVENTS Practice: Roll a die and flip a coin: 1-P(heads and 6) = ½ x 1/6 = 2-P(tails and a 5) =

12-4: INDEPENDENT EVENTS Practice: Roll a die and flip a coin: 1-P(heads and 6) = ½ x 1/6 = 1/12 2-P(tails and a 5) =

12-4: INDEPENDENT EVENTS Practice: Roll a die and flip a coin: 1-P(heads and 6) = ½ x 1/6 = 1/12 2-P(tails and a 5) = ½ x 1/6 =

12-4: INDEPENDENT EVENTS Practice: Roll a die and flip a coin: 1-P(heads and 6) = ½ x 1/6 = 1/12 2-P(tails and a 5) = ½ x 1/6 = 1/12

12-4: Independent Events Example: Suppose you spin each of these two spinners. What is the probability of spinning an even number and a vowel? P(even) = P(vowel) = P(even and vowel) = S T R O P 1 2 3 6 5 4 Slide 35

12-4: Independent Events Example: Suppose you spin each of these two spinners. What is the probability of spinning an even number and a vowel? P(even) = (3 evens out of 6 outcomes) P(vowel) = (1 vowel out of 5 outcomes) P(even and vowel) = S T R O P 1 2 3 6 5 4 Slide 36

12-4: Independent Events Example: Suppose you spin each of these two spinners. What is the probability of spinning an even number and a vowel? P(even) = (3 evens out of 6 outcomes) P(vowel) = (1 vowel out of 5 outcomes) P(even and vowel) = S T R O P 1 2 3 6 5 4 Slide 37

12-4: Independent Events Find the probability P(jack and factor of 12)
x = Slide 38

12-4: Independent Events Find the probability 1 5 5 8
P(jack and factor of 12) x = Slide 39

P(jack and factor of 12) 5 40 x = 1 8 Slide 40

12-4: Independent Events Find the probability P(6 and not 5) x =
Slide 41

12-4: Independent Events Find the probability P(6 and not 5) 1 6 5 6 x
= Slide 42

12-4: Independent Events Find the probability P(6 and not 5) 1 6 5 6 5
36 x = Slide 43

12-4: Dependent Event P(A∩B) = P(A) ● P(B)
What happens during the second event depends upon what happened before. In other words, the result of the second event will change because of what happened first. Determining the probability of a dependent event is usually more complicated than finding the probability of a independent event. The probability of two dependent events, A and B, is equal to the probability of event A times the probability of event B. However, the probability of event B now depends on event A. P(A∩B) = P(A) ● P(B) Slide 44

12-4: P(A and B) = P(A) x P(B )
So for both INDEPENDENT & DEPENDENT EVENTS… To find the probability that two events, A and B will occur in sequence, multiply the probability A occurs by the conditional probability B occurs, given A has occurred. P(A and B) = P(A) x P(B ) ***EXCEPTION: For dependent, P(B given A) DEPENDENT EXAMPLE: Two cars are selected from a production line of 12 where 5 are defective. Find the probability both cars are defective. A = first car is defective B = second car is defective. P(A) = 5/12 P(B given A) = 4/11 P(A and B) = 5/12 x 4/11 = 5/33 =

12-4: Probability Examples
Ex.1) Independent Events: Spinner #1 is partitioned into three equal sections, colored black, white, and grey. Spinner #2 is partitioned into four equal sections, colored red, blue, green, and yellow. If both spinners are spun, what is the probability of getting black and red?

12-4: Probability Example
Since we expect to get black one-third of the time, and we expect to get red one-quarter of the time, then we expect to get black one-third and red one-quarter of the time. . . Imagine a tree diagram where the first column shows the three outcomes for Spinner #1, each of which is followed by the four outcomes for Spinner #2 in the second column. Three groups of four branches creates 12 possible outcomes.

Imagine a tree diagram where the first column shows the three outcomes for Spinner #1, each of which is followed by the four outcomes for Spinner #2 in the second column. Three groups of four branches creates 12 possible outcomes.

Ex.2) Dependent Events: A bag contains 10 marbles; 5 red, 3 blue, and 2 silver. If you draw one marble at random and hold it in your left hand, and then draw a second marble at random and hold it in your right hand, what is the probability that you are holding two silver marbles? It’s easy to determine the probability of the first marble being silver. However, notice that if you start by getting a silver marble and then try for the second, the bag will be different. How? Now, there is only one silver marble in a bag containing a total of 9 marbles. . .

12-4: Dependent Event Example: There are 6 black pens and 8 blue pens in a jar. If you take a pen without looking and then take another pen without replacing the first, what is the probability that you will get 2 black pens? P(black first) = P(black second) = Slide 53

12-4: Dependent Event Example: There are 6 black pens and 8 blue pens in a jar. If you take a pen without looking and then take another pen without replacing the first, what is the probability that you will get 2 black pens? P(black first) = P(black second) = (There are 13 pens left and 5 are black) THEREFORE……………………………………………… P(black and black) =?? Slide 54

12-4: Dependent Event Example: There are 6 black pens and 8 blue pens in a jar. If you take a pen without looking and then take another pen without replacing the first, what is the probability that you will get 2 black pens? P(black first) = P(black second) = (There are 13 pens left and 5 are black) THEREFORE……………………………………………… P(black and black) = Slide 55

Find the probability 12-4: Dependent Events P(Q∩Q)
All the letters of the alphabet are in the bag 1 time. Do not replace the letter. ∩ means “and”. “AND” MEANS MULTIPLY!!! x = Slide 56

Find the probability 12-4: Dependent Events 1 26 25 P(Q∩Q)
All the letters of the alphabet are in the bag 1 time. Do not replace the letter. ∩ means “and”. “AND” MEANS MULTIPLY!!! 1 26 25 x = Slide 57

Find the probability 12-4: Dependent Events P(Q∩Q)
All the letters of the alphabet are in the bag 1 time. Do not replace the letter. ∩ means “and”. “AND” MEANS MULTIPLY!!! 1 26 25 650 x = Slide 58

12-4: Examples

Find the probability of rolling a 4.
Finding Probabilities of Events You roll a six-sided die whose sides are numbered from 1 through 6. Find the probability of rolling a 4. SOLUTION Only one outcome corresponds to rolling a 4. number of ways to roll a 4 = P (rolling a 4) = number of ways to roll the die

Find the probability of rolling a 4.
Finding Probabilities of Events You roll a six-sided die whose sides are numbered from 1 through 6. Find the probability of rolling a 4. SOLUTION Only one outcome corresponds to rolling a 4. number of ways to roll a 4 1 6 = P (rolling a 4) = number of ways to roll the die

Find the probability of rolling an odd number.
Finding Probabilities of Events You roll a six-sided die whose sides are numbered from 1 through 6. Find the probability of rolling an odd number. SOLUTION Three outcomes correspond to rolling an odd number: rolling a 1, 3, or a 5. =?? number of ways to roll an odd number P (rolling odd number) = number of ways to roll the die

Find the probability of rolling an odd number.
Finding Probabilities of Events You roll a six-sided die whose sides are numbered from 1 through 6. Find the probability of rolling an odd number. SOLUTION Three outcomes correspond to rolling an odd number: rolling a 1, 3, or a 5. number of ways to roll an odd number 3 6 1 2 = = P (rolling odd number) = number of ways to roll the die

Find the probability of rolling a number less than 7.
Finding Probabilities of Events You roll a six-sided die whose sides are numbered from 1 through 6. Find the probability of rolling a number less than 7. SOLUTION: All six outcomes correspond to rolling a number less than 7. number of ways to roll less than 7 P (rolling less than 7 ) = =?? number of ways to roll the die

Find the probability of rolling a number less than 7.
Finding Probabilities of Events You roll a six-sided die whose sides are numbered from 1 through 6. Find the probability of rolling a number less than 7. SOLUTION: All six outcomes correspond to rolling a number less than 7. number of ways to roll less than 7 6 = = 1 P (rolling less than 7 ) = number of ways to roll the die

12-4:SUMMARY: Independent & Dependent 1-Independent & Dependent Events P(A∩B) = P(A) ● P(B) P(A∩B) means probability of event A and event B occurring. Multiply! Symbol ∩ means the same as the word “and”….both means multiply.

P(jack and factor of 12) 5 40 x = 1 8 Slide 67

12-4: Dependent Events Find the probability P(Q∩Q)
All the letters of the alphabet are in the bag 1 time. Do not replace the letter. ∩ means “and”. “AND” MEANS MULTIPLY!!! Skills Practice Workbook pg. 82 (1-15) 1 26 25 650 x = Slide 68

12-5: Adding Probabilities
TWO OR MORE EVENTS 12-5: Adding Probabilities Wednesday Definitions: Simple event-consists of only one event Compound event-consists of two or more simple events Mutually exclusive events-cannot occur at the same time Inclusive events (not mutually exclusive)-can occur at the same time

Compare “A and B” to “A or B” Multiplication vs. Adding
The compound event “A and B” means that A and B both occur in the same trial. Use the multiplication rule to find P(A and B). The compound event “A or B” means either A can occur without B, B can occur without A or both A and B can occur. Use the addition rule to find P(A or B). A B A B A or B A and B

Mutually Exclusive Events
Mutually Exclusive events-two events that cannot occur at the same time. P(A U B)=P(A) + P(B) U means “or”. “OR” MEANS ADD!!! Ex. Probability of drawing a 2 or an ace is found by adding their individual probabilities. P(2 or ace)= P(2) + P(ace) *Add probabilities. *There are 4 two’s and 4 aces in a deck. 8 = *Simplify. The probability of drawing a 2 or an ace is 2/13.

Redundancy-meaning subtract what both have in common!!
Inclusive Events Inclusive events (not mutually exclusive)-can occur at the same time. P(A U B)=P(A) + P(B) – P(A and B) - Redundancy Redundancy-meaning subtract what both have in common!! Ex. Drawing a spade or drawing an ace Probability of drawing a spade: 13 outcomes, so 13/52 = 1/4 Probability of drawing an ace: 4 outcomes, so 4/52 = 1/13 Ace of spades is common to both events, probability is 1/52 So probability of drawing a spade or an ace is 1/4 + 1/13 – 1/52 = 16/52 = 4/13

Mutually Exclusive Events
Two events, A and B, are mutually exclusive if they cannot occur in the same trial. EXAMPLES: A = A person is under 21 years old B = A person is running for the U.S. Senate A = A person was born in Philadelphia B = A person was born in Houston Mutually exclusive… P(A and B) = 0 *No Redundancy No common ground A B When event A occurs it excludes event B in the same trial.

When events are mutually exclusive,
12-5: The Addition Rule EXAMPLES: A card is drawn from a deck. Find the probability the card is a king or a 10. A = the card is a king B = the card is a 10. REDUNDANCY! P(A) = 4/52 AND P(B) = 4/52 AND P(A and B)= 0/52 P(A or B) = 4/52 + 4/52 – 0/ = 8/52 P(A) + P(B) – P(A and B) When events are mutually exclusive, P(A or B) = P(A) + P(B)

Non-Mutually Exclusive Events or
INCLUSIVE If two events can occur in the same trial, they are non-mutually exclusive (inclusive). EXAMPLES: A = A person is under 25 years old B = A person is an attorney A = A person was born in Philadelphia B = A person watches Walking Dead on TV A and B Non-mutually exclusive P(A and B) ≠ 0 A B

12-5 : INCLUSIVE EVENTS - Redundancy EXAMPLES:
The probability that one or the other of two events will occur is: P(A) + P(B) – P(A and B) - Redundancy EXAMPLES: A card is drawn from a deck. Find the probability it is a king or it is red. A = the card is a king B = the card is red. P(A) = 4/52 or P(B) = 26/52 but P(A and B) = 2/52 P(A or B) = 4/ /52 – 2/52 = 28/52

12-5: Examples 1. A die is rolled. Find the probability of rolling a 6 or an odd number. -are the events mutually exclusive? -find P(A), P(B) and, if necessary, P(A and B) -use the addition rule to find the probability

12-5: Examples 1. A die is rolled. Find the probability of rolling a 6 or an odd number. -are the events mutually exclusive (is a 6 an odd number)? -find P(A), P(B) and, if necessary, P(A and B)meaning if you have redundancies… -use the addition rule to find the probability

12-5: Examples 1. A die is rolled. Find the probability of rolling a 6 or an odd number. -are the events mutually exclusive (is a 6 an odd number)? NO -find P(A), P(B) and, if necessary, P(A and B) Redundancies? NO 1/6 + 3/6 =

12-5: Examples 1. A die is rolled. Find the probability of rolling a 6 or an odd number. -are the events mutually exclusive (is a 6 an odd number)? NO -find P(A), P(B) and, if necessary, P(A and B) Redundancies? NO 1/6 + 3/6 = 4/6 = 2/3

12-5: Example 2. A card is selected from a standard deck. Find the probability that the card is a face card or a heart. How many face cards in a deck? 12 How many heart cards in a deck? 13 -are there any redundancies (heart & face card)? -find P(A), P(B) and, if necessary, P(A and B) -use the addition rule to find the probability

12-5: Example 2. A card is selected from a standard deck. Find the probability that the card is a face card or a heart. How many face cards in a deck? 12 How many heart cards in a deck? 13 -are there any redundancies (heart & face card)? YES SO… -find P(A), P(B) and, if necessary, P(A and B)

12-5: Example 2. A card is selected from a standard deck. Find the probability that the card is a face card or a heart. How many face cards in a deck? 12 How many heart cards in a deck? 13 -find P(A) + P(B) - P(A and B) -REDUNDANCIES 12/ /52 – 3/52 =

12-5: Example 2. A card is selected from a standard deck. Find the probability that the card is a face card or a heart. How many face cards in a deck? 12 How many heart cards in a deck? 13 -find P(A) + P(B) - P(A and B) -REDUNDANCIES 12/ /52 – 3/52 = 22/52

12-5: Contingency Table Mark conducted a survey asking adults in three different cities if they liked a new juice that just came out. The results were as follows: Omaha Seattle Miami Total Yes 100 150 150 400 No 125 130 95 350 Undecided 75 170 5 250 Total 300 450 250 1000 One of the responses is selected at random. Find: 1. P(Miami and Yes) 2. P(Miami and Seattle) 3. P(Miami or Yes) 4. P(Miami or Seattle)

12-5: Contingency Table 1. P(Miami and Yes) 2. P(Miami and Seattle)
Omaha Seattle Miami Total Yes 100 150 150 400 No 125 130 95 350 Undecided 75 170 5 250 Total 300 450 250 1000 One of the responses is selected at random. Find: 1. P(Miami and Yes) 2. P(Miami and Seattle) = 250/1000 • 150/250 = 150/ = 3/20 = 0

12-5: Contingency Table 3. P(Miami or Yes) 4. P(Miami or Seattle)
Omaha Seattle Miami Total Yes 100 150 150 400 No 125 130 95 350 Undecided 75 170 5 250 Total 300 450 250 1000 3. P(Miami or Yes) 4. P(Miami or Seattle) =250/ /1000 – 150/1000 = 500/1000 = 1/2 =250/ /1000 – 0/1000 = 700/1000 = 7/10

P(A or B) = P(A) + P(B) - P(A and B)
Summary Probability at least one of two events occur P(A or B) = P(A) + P(B) - P(A and B) Add the simple probabilities, BUT to prevent double counting, don’t forget to subtract the probability of both occurring… THE REDUNDANCY. ***Skills Practice Worksheet-Adding Probabilities (#1-20).

Thursday Review: TWO OR MORE EVENTS: (mutually exclusive and not mutually exclusive) (1) Mutually Exclusive events-two events that cannot occur at the same time. Exclusive-excluding or restricted P(A U B)=P(A) + P(B) U means “or”. “OR” MEANS ADD!!!

TWO OR MORE EVENTS: (mutually exclusive and inclusive events) (2) not mutually exclusive-or (inclusive events) can occur at the same time. Inclusive-Including all items normally expected or required, containing (a specified element) as part of a whole. P(A U B)=P(A) + P(B) – P(A and B) U means “or”. “OR” MEANS ADD!!!

Conditional Probability 3-P(B│A) means probability of event B given event A occurred. Symbol “│” tells you which one we want… it means given.

Conditional Probability
Probability of second event occurring given first event has occurred Drawing a spade from a deck given you have previously drawn the ace of spade After drawing ace of spades have 51 cards left. Remaining cards now include only 12 spades. Conditional probability is then 12/51. P(A│B) means probability of event A given event B occurred!!!

PRACTICE

Probability Practice Problems
Suppose you have a bag of marbles numbered 1 – 15. P(even)?

Suppose you have a bag of marbles numbered 1 – 15. P(even)? even numbers = 2, 4, 6, 8, 10, 12, 14 = 7 total numbers through =15

Suppose you have a bowl of cds numbered 1 – 15. P(even, more than 10)? A COMMON SENSE PROBLEM!!! THINK ABOUT IT!!!

Suppose you have a bowl of cds numbered 1 – 15. P(even, more than 10)? Even #’s & #’s greater than 10 =12, 14= 2 Total numbers = 15 = 15

Suppose you have a bowl of cards numbered 1 – 15. P(even or more than 10)? The “or” indicates the card must be even or more than 10. You must be careful not to include a number twice

Suppose you have a bowl of cards numbered 1 – 15. P(even or more than 10)? even #’s = 2, 4, 6, 8, 10, 12, 14 = 7/15 #’s greater than 10 = 11, 12, 13, 14, 15 = 5/15 Since 12 and 14 are common to both sets, you will subtract 2/15 Solution: 7/15 + 5/15 – 2/15 = 10/15 = 2/3

Suppose you have a bowl of coins numbered 1 – 15. A coin is drawn, replaced, and a second coin is drawn. P(even and even)?

Suppose you have a bowl of coins numbered 1 – 15. A coin is drawn, replaced, and a second coin is drawn. P(even and even)? Find the probability of each independent event and multiply Even #’s on first draw = 2, 4, 6, 8, 10, 12, 14 = 7/15 Even #’s on second draw = 2, 4, 6, 8, 10, 12, 14 = 7/15 Solution: 7/15 x 7/15 = 49/225

Suppose you have a bowl of disks numbered 1 – 15. A disk is drawn, not replaced, and a second disk is drawn. P(even and even)?

Suppose you have a bowl of disks numbered 1 – 15. A disk is drawn, not replaced, and a second disk is drawn. P(even and even) Find the probability of each independent event and multiply even #’s on first draw = 2, 4, 6, 8, 10, 12, 14 = 7/15 Even #’s on second draw = one less even number than previous set/one less disk from bowl = 6/14 Solution: 7/15 x 6/14 = 42/210 = 1/5 12-4 Enrichment Worksheet (#1-10)

CHAPTER 12: PROBABILITY AND STATISTICS

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