Lesson 15.1: Intro to Probability

Learning Goals:

1) What is sample space and how do we determine the sample space of a given situation?

2) What is probability?3) What is the complement of a probability?

Do Now: Answer these 7th grade math question in order to prepare for probability.

1. State the probabilities of each event given below:

a. Flipping a fair coin, with sides labeled heads and tails, and the coin landing on tails.

P (T )=12=50 %

b. Choosing a marble out of a bag, with nine blue marbles and one red marble, and

the marble being red. P (R )= 110

=10 %

c. Rolling a fair number cube, with faces labeled one to six, and the cube landing on a

number less than six. P (¿<6 )=56=83 %

d. Spinning the arrow on a spinner, with four equal sectors labeled one to four, and the

arrow landing on a number greater than one. P (¿>1 )=34=75 %

Which event above is most likely to occur? Justify your answer. Event C because it has the highest probability.

2. A number cube has faces numbered 1 through 6, and a coin has two sides, heads and tails. The number cube will be rolled once, and the coin will be flipped once.

Find the probabilities of the following events:

a. The number cube shows a 6, and the coin shows heads.

P (6 )∧P (H )=16∙ 12= 1

12

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b. The number cube shows a 6, or the coin shows heads.

P (6 )∨P (H )=16+ 1

2=4

6=2

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What keywords in pards 2a and 2b let you know how to answer each question?

“and” → multiply probabilities

“or” → add probabilities

3. Which of the following could NOT represent the probability of an event?

(a) 0 (b) 1 (c) 0.75 (d) −0.75 probability of an event 0≤P (E)≤1

Introduction to Probability

Mathematics seeks to quantify and model just about everything. One of the greatest challenges is to try to quantify chance. However, that is exactly what probability seeks to do.

With probability, we attempt to assign a number to how likely an event is to occur.

Terminology in probability is important, so we introduce some basic terms here:

The probability of any event is a number that lies between 0 and 1, inclusive.2

BASIC PROBABILITY TERMINOLOGY

1. Experiment: Some process that occurs with well-defined outcomes. Flip a coin

2. Outcome: A result from a single trial of the experiment. Heads or Tails

3. Event: A collection of one or more outcomes. Flip coins once (H T) or twice (HT TH HH TT).

4. Sample Space: A collection of all of the outcomes of an experiment.

5. Probability: The measure of how likely an event is to occur.

P (E )= n(E)n(S)

= number of outcomes that fall into the event Enumber of outcomes that fall into the sample space

= PartWhole

An impossible event has a 0 % chance of happening (probability of 0). A certain event has a 100 % chance of happening (probability of 1).

The number line below shows how we could interpret different probabilities:

Probability can be expressed in many forms. The most popular are: fractions, decimals, and percents.

What is the probability of spinning a 5 on an octagon with 8

equal sections? P (5 )=18=0.125=12.5 %

Sample Spaces

Sample Space: A collection of all the outcomes of an experiment. When listing a sample space list to represent the outcomes, you must ask

yourself “how many different ways can this happen?” When creating a sample space, there are a variety of ways to

illustrate/represent the outcomes. Lists and tree diagrams are two examples of how to create a sample

space.

Directions: List the possible outcomes for the following events by creating a list or tree diagram.

a. Tossing two coins 2 outcomes (H, T) = 2 branches!

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b. Drawing three marbles from a bag containing many red and yellow marbles.

2 outcomes (R, Y) = 2 branches!

How many outcomes are at least 2 Reds? (2 or 3 Reds = 4 outcomes)

c*. The gender of three children

2 outcomes (B, G) = 2 branches!

Complementary Events

The sum of the probabilities of all possible outcomes of a trial must be equal to 1.

a. Sum of probabilities of flipping a coin. H or T→ 12+1

2=1

b. Sum of probabilities of rolling a six-sided die. 1 or 2 or 3 or 4 or 5 or 6

16+ 1

6+ 1

6+ 1

6+ 1

6+ 1

6=1

The complement of any event (event A) is all outcome that are NOT event A.

a. What is the complement of flipping a coin and getting tails? Not getting tails.

b. What is the complement of rolling a six-sided die and rolling a 6? NOT getting a 6

How do we find the probability of the complement of any event (event A)?

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P (not A )=1−P(A)

a. Suppose that the probability that a particular flight is on time is 0.78. What is the probability that the flight is not on time? P (NOT on time )=1−.78=.22

b. The probability that there will be no rain on Thursday is 56 . What is the probability

that it will rain on Thursday? P (Rain )=1−56=1

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c. If a marble is randomly chosen from a bag that contains exactly 8 red marbles, 6 blue marbles, and 6 white marbles, what is the probability that the marble will not be white? Total = 20

P (W )= 620 8+6

20=14

20= 7

10

P (NOTW )=1−P(W )

¿1− 620

¿ 1420

= 710

Probability with “AND” / “OR”

If an event uses key words such as “and” / “or”, you must use a specific operation between your probabilities.

If an event uses “and”, then you must multiply the probabilities. If an event uses “or”, then you must add the probabilities.

a. A spinner is created with four equal sized sectors as shown below. An experiment is run where the spinner is spun twice and the outcome is recorded each time. What is the probability of spinning a 4 and then spinning a 2?

14∙ 1

4= 1

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b. A basket of fruit contains 10 apples, 6 bananas, and 4 oranges. A fruit is selected at random. Find the probability you pick an apple or an orange.

1020

+ 420

=1420

= 710

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Homework 15.1: Introduction to Probability

1. Which inequality represents the probability, x, of any event happening?

(1) x≥0 (2) 0<x<1 (3) x<1 (4) 0≤ x≤1

2. Mary chooses an integer at random from 1 to 6. What is the probability that the integer she chooses is a prime number?

(1) 56 (2) 3

6 (3) 26 (4) 4

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3. A bag contains eight green marbles, five white marbles, and two red marbles. What is the probability of drawing a red marble from the bag?

(1) 115 (2) 2

15 (3) 213 (4) 13

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4. A six-sided number cube has faces with the numbers 1 through6 marked on it. What is the probability that a number less than 3 will occur on one toss of the number cube?

(1) 16 (2) 2

6 (3) 36 (4) 4

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5. As captain of his football team, Jamal gets to call heads or tails of a fair coin at the beginning of each game. At the last three games, the coin has landed with heads up. What is the probability that the coin will land with heads up at the next game? Explain your answer.

6. Marilyn selects a piece of candy at random from a jar that contains four peppermint, five cherry, three butterscotch, and two lemon candies. What is the probability that the

candy she selects is not a cherry candy? (1) 0 (2) 514 (3) 9

14 (4) 1414

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7. Which event is certain to happen?

(1) Everyone walking into a room will have red hair.

(1) All babies born in June will be males.

(3) The Yankees baseball team will win the World Series.

(4) The Sun will rise in the east.

8. The faces of a cube are numbered from 1 to 6. If the cube is rolled once, which outcome is least likely to occur?

(1) rolling an odd number (2) rolling an even number

(3) rolling a number less than 6 (4) rolling a number greater than 4

9. A blood collection agency test 50 blood samples to see what type they are. Their results are shown in the table below.

If a blood sample is picked at random, what is the probability, as a percent, it will be type B?

If a blood sample is picked at random, what is the probability, as a percent, it will not be type O?

10. A box contains five blue, four red, and three green tickets. Once ticket is randomly selected from the box. What is the probability that the ticket is blue or green?

11. Carl owns a car and a motorcyle. His car will only start 80 % of the time and his motorcyle will only start 60 % of the time. What is the proability that both his car and motorcycle will start?

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Lesson 15.2: Addition Law of Probability

Learning Goals:

1) What is the addition law of probability?2) How do we use the addition law to find the probability of combined events?

Warm-up: We want to find the probability that a person chosen at random from the group below is wearing a blue shirt OR a tie.

Use the picture to answer the following:

Find the probability by counting up the number of people who fall into this

category. 12 total, 3 blue shirts, and 2 ties 312

+ 212

= 512

Find the probability by using what we learned yesterday about “or” probability. What do you notice? One person counted twice Why does this happen? How can we adjust the “or” rule from yesterdays lesson to take this into account?

P (B )+P (T )−P (B∩T )=P (B∪T )3

12+ 2

12− 1

12= 4

12

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Addition Rule of Probability

P (A∪B )=P ( A )+P (B )−P ( A∩B )

∩ means AND

∪ means OR

1. Use the spinner to the right to answer the following questions:

What is the probability that you spin the spinner and it lands on

green or an odd number? 28+ 4

8−1

8=5

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What is the probability that you spin the spinner and it lands on red or a prime

number? 38+ 4

8−1

8=6

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Using the formula without context:

2. Let P (A )=0.6 ,P (B )=0.5 ,∧P ( A∩B )=0.2. What is P(A∪B)?

P (A∪B )=P ( A )+P (B )−P ( A∩B )=.5+.5−.2=.9

3. Given P (A∪B )=0.82 ,P ( A )=0.45 ,∧P (B )=0.57 , find P(A∩B).

P (A∪B )=P ( A )+P (B )−P ( A∩B )

.82=.45+.57−P(A∩B)

.82=1.02−P(A∩B)

−0.2=−P(A∩B)

P (A∩B )=0.2

Using the formula with context:

4. In the real estate ads, 64 % of homes have garages, 21 % have swimming pools, and 68 % have garages or swimming pools. What’s the probability a home has both?

P (A )=.64 P (A∪B )=P ( A )+P (B )−P ( A∩B )

P (B )=.21 .68=.64+.21−P(A∩B)

P (A∪B )=.68 .68=.85−P(A∩B)

−.17=−P (A∩B )

P (A∩B )=.17

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5. After a recent disaster, 200 people in a community were asked what kind of help they gave to the victims. 65 said they donated food, 50 people said they donated money, and 30 people said they dontated both. What is the probability that a person selected at random from the sample donated food or money?

Grand Total ¿200 65200

+ 50200

− 30200

= 85200

=1740

=.425

P (A )= 65200

P (B )= 50200

P (A∩B )= 30200

6. Let P (A )=0.6 ,P (B )=0.7∧P (A∪B )=0.8. Determine P (A ∩B ) .

7. A student figured out that his probability of passing the anthropology final was 80 %. Based off his performance so far he determined the probability of passing his biology final was 85% and that passing them both was 70%. What is the probability that he will pass either his anthropology or biology final?

8. Let P (A )=0.23 , P (B )=0.84 ,∧P (A∧B )=0.41. What is P(A∨B)?

9. Harrison High School has a population of 1205 students. The number of students who participate in band is 85. The number of students who participate in chorus is 125.

If the probability that a student participates in either band or chorus is 1521205 , what is the

probability that a student participates in both band and chorus?

10. Based on the spinner to the right, find:

a) The probability that the spinner lands on a yellow or a multiple of 2.

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b) The probability that the spinner lands on a blue or a number less than 4

Homework 15.2: Addition Law of Probability

1. Given P (A∪B )=0.73 ,P ( A )=0.35 ,∧P (B )=0.51 . Find P(A∩B)

2. On New Year’s Eve, the probability of a person having a car accident is 0.29. The probability of a person driving while intoxicated is 0.32 and the probability of a person having a car accident or being intoxicated is 0.46. What is the probability of a person having a car accident while intoxicated?

3. Given P (A ∩B )=0.11 , P (A )=0.35 ,∧P (B )=0.51. Find P(A∪B).

4. There are 984 students at a high school. 207 students take photography and 312 students take computer science. The probability that a student takes photography or

computer science is 431984 . Determine the probability that a student takes both

photography and computer science.

5. A box is filled with candies in different colors. We have 40 white candies, 24 green ones, 12 red ones, 24 yellow ones, and 20 blue ones. If we have selected one candy from the box without peeking into it, find the probability of getting a green or red candy.

6. We have numbered cards from 1¿20 and picked one at random. Find the probability that the card picked is numbered a multiple of 2∨5.

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Lesson 15.3: Probability with Two-Way Tables

Learning Goals:

1) What is a two-way table and how can we use one to determine the probability of an event occurring?

2) How do we determine the probability of overlapping events?

Introduction to Two-Way Tables

What is a two-way table? A two-way table shows data that pertain to two different categories. The data from one sample group is shown as it relates to two different categories.

Example 1: The data from a survey of 50 students is shown in the two-way table below. The students were asked whether or not they were taking a foreign language and whether or not they played a sport.

The totals shown are for the corresponding row or column with a grand total of 50 students in the data set.

(a) How many students play a sport? 24(b) How many students do not take a foreign language? 13(c) How many students play a sport and do not take a foreign language? 10(d) How many students play a sport or do not take a foreign language?

24+13−10=27

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Practice: 80 Grade 12 students each study one Science. The table shows some information about these students.

(a) Complete the table.(b) How many students do not study Chemistry? 26+33=59(c) How many students are female and study Physics? 14(d) How many students are female or study Physics? 47+33−14=66

Constructing Two-Way Tables

How do we create a two-way table from a word problem? Use the following steps:

1. Create a table using the two categories.

2. Use the values given to fill in the table.

3. Use reasoning to complete the table. Remember, the totals are for each row and column.

o The column labeled “total” should have the same sum as the row labeled “total”

Example 2: Felipe surveyed students at his school. He found that 78 students own a cell phone and 57 of those students own an MP3 player. There are 13 students that

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do not own a cell phone, but own an MP3 player. Nine students do not own either device. Construct a two-way table summarizing the data.

a) How many students do not own a cell phone? 22

b) How many students own both a cell phone and an MP3 player? 57

c) How many students own a cell phone or an MP3 player? 78+70−57=91

Example 3: There are 150 children at summer camp and 71 signed up for swimming. There were a total of 62 children that signed up for canoeing and 28 of them also signed up for swimming. Construct a two-way table summarizing the data. What is the probability that a child signed up for swimming or canoeing?

Probability with Two-Way Tables

How can we use a two-way table to find the probability of a given event?

RECALL: Probability ¿ outcome you wanttotal possibleoutcomes

What is the addition law of probability? For any two events A and B, the addition law of probability states that:

P (either A∨B )=P ( A )+P (B )−P(A∧B)⏟overlap

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Example 4: The table below shows the number of left and right handed tennis players in a sample of 50 males and females.

If a tennis player was selected at random from the group, find the probability that the player is

(a) male and left-handed 350

(b) right handed 4550

(c) female or right handed 1850

+ 4550

−1650

=4750

Practice 1: The data in the table below refers to a sample of 60 randomly chosen plants.

A plant is chosen at random from the above group.

(a) Find the probability of a plant being in a shady environment. 3260

(b) Find the probability of a plant having a low growth rate and being in a dark

environment. 860

(c) Find the probability of a plant not being in a dark environment. 4960

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(d) Find the probability of a plant having a high growth rate or being in a dark

environment. 2560

+ 1160

− 360

=3360

Practice 2: In a survey of 52 students it was found that 30 study Spanish and 15 have computers. Seven of the students who study Spanish also have computers.

(a) Based on the information given above, complete this table.

A student is selected at random to attend a computer workshop given in Spanish.

(b) What is the probability that the student:

(i) has a computer and studies Spanish? 752

(ii) has a computer but does not study Spanish? 852

(iii) does not have a computer or does not study Spanish? 3752

+ 2252

−1452

=4552

Example: When an animal is selected at random from those at a zoo, the probability that it is North American (meaning that its natural habitat is in the North American continent) is 0.65, the probability that it is both North American and a carnivore is 0.16, and the probability that it is neither American nor a carnivore is 0.17. Complete the table below showing the probabilities of the events corresponding to the cells of the table.

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Practice: When an avocado is selected at random from those delivered to a food store, the probability that it is ripe is 0.12, the probability that it is bruised is 0.054, and the probability that it is ripe and bruised is 0.019.

(a) Complete the table below showing the probabilities of the events corresponding to the cells of the table.

(b) Find the probability that an avocado randomly selected from those delivered to the store is:

i. Not bruised .946

ii. Ripe and bruised .019

iii. Not ripe or bruised .88+.054−.035= .899

(c) Extend Your Thinking: Find the probability, to the nearest thousandth, that an avocado randomly selected is bruised, given that it is ripe.

Only look at ripe .019.12

=.158

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Homework 15.3: Probability with Two-Way Tables

1. The table shows the number of girls and boys who can swim.

One person is chosen at random. Write down the probability that:

a) the person is a boy

b) the person can swim

c) the person is a boy and can swim

d) the person is not a boy or cannot swim

2. Complete the table for the activities chosen by 74 teenagers on an activity holiday.

One teenager is chosen at random. Write down the probability that:

a) the teenager does rock climbing

b) the teenager is not a boy

c) the teen ager is not a boy and does rock climbing

d) the teenager is a girl or does mountain climbing

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3. Peter throws a dice and flips a coin 150 times as part of an experiment. He records 71 heads and a six 21 times. On 68 occasions, he gets neither a head nor a six. Complete the table below using this information.

One teenager is chosen at random. Write down the probability of flipping heads on the coin, given that Peter has already rolled a 6.

4. Obedience School for Dogs is a small franchise that offers obedience classes for dogs. Some people think that larger dogs are easier to train and, therefore, should not be charged as much for the classes. To investigate this claim, dogs enrolled in the classes were classified as large (30 pounds or more) or small (under 30 pounds). The dogs were also classified by whether or not they passed the obedience class offered by the franchise. 45% of the dogs involved in the classes were large. 60% of the dogs passed the class. Records indicate that 40% of the dogs in the classes were small and passed the course. Complete the following two-way table using the probabilities given above.

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Lesson 15.4: Proving Independence and Disjoint Events

Learning Goals:

1) How do we prove that two events are independent?2) How do show that two events are disjoint (or mutually exclusive)?

Warm-Up

a. The probability of someone having brown eyes is 55%. If two different people are chosen, what is the probability that both people have brown eyes? This is an example of independent events! 0.55×0.55=.3025

b. A standard die has the numbers 1 – 6 on its sides. A spinner is divided into 4 equal sections (green, blue, red, yellow). What is the probability that you roll a 5 or that you land on green when you spin the spinner? This is an example of disjoint events. There is no intersection or overlapping! So it is mutually exclusive.

P (S ) orP (Green)=16+ 1

4= 5

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1. There is a 34% chance that a person picked at random from the adult population is a regular smoker of cigarettes and an 18% chance that a person picked has emphysema. If the percent of the adult population that are both regular smokers and suffer from emphysema is 14%. Is being a smoker independent of having emphysema? Justify your answer.

A=being asmoker∧B=emphysema P (A )=0.34

P (A∧B )=P (A )∙ P(B) P (B )=0.18

0.14?¿

0.34 ∙0.18 P (A∧B )=0.14

0.14≠0.0612

Not independent (or dependent) because the values of the test are not equal!

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Product Test for Independence (must memorize)

If two events are independent, then P (A∧B )=P (A )∙ P(B)

Can also be written as P (A ∩B )=P(A)∙ P ¿

One event does not affect the other!

2. The two-way frequency table below shows the proportions of a population that have given hair and eye color combinations. Show that the events of having green eyes and red hair are dependent.

A=green eyes∧B=redhair P (A )=0.25

P (A∧B )=P (A )∙ P(B) P (B )=0.18

0.15?¿

0.25 ∙0.18 P (A∧B )=0.15

0.15≠0.045

Not independent (or dependent) because the values of the test are not equal!

3. The table below shows the results of a survey in which young adults ages 18 – 24 were asked if they were ever used Instashop (Yes or No). Is the age of young adults independent of their use of Instashop? Show the mathematics that led to your answer.

A=female∧B=used Instashop P (A )=270510

P (A∧B )=P (A )∙ P(B) P (B )=388510

216510

?¿

270510

∙ 388510 P (A∧B )=216

510

0.4235≠0.4028

Gender and Instashop are not independent (or dependent)

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4. A deck of 40 cards consists of the following:

10 black cards showing squares, numbered 1 – 10 10 black cards showing circles, numbered 1 -10 10 red cards showing X’s, numbered 1 – 10 10 red cards showing diamonds, numbered 1 – 10

A card will be selected at random from the deck.

a. Are the events “the card shows a square” and “the card is red” disjoint? Explain. Disjoint events because there are no red cards showing a square! (no intersection)

P (A∧B )=0

b. Calculate the probability that the card will show a square or will be red.

1040

+ 2040

=3040

=34

c. Are the events “the card shows a 5” and “the card is red” disjoint? Explain. Not disjoint events because there are two red cards that show a 5!

d. Calculate the probability that the card will show a 5 or will be red.

P (S∨R )=P (S )+P (R )−P (S∧R )= 440

+ 2040

− 240

=2240

=1120

5. A set of 40 cards consists of the following:

10 black cards showing squares 10 black cards showing circles

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Disjoint Events (Mutually Exclusive)

If A∧B are disjoint, then P (A∧B )=0 or P (A∩B )=0

So, the addition rule for disjoint events can be written as

P (A∨B )=P ( A )+P (B) (must memorize)

10 red cards showing X’s 10 red cards showing diamonds

A card will be selected at random from the set.

a. Are the events “shows black” and “shows a diamond” disjoint events? Explain how you know. Yes, because the two events do not overlap.

b. Find the probability that the card is black or shows a diamond. 2040

+ 1040

−0=3040

6. Given two events A∧B where P (A )=0.3 , P (B )=0.6 ,∧P ( A∨B )=0.7, determine if the two events are disjoint. Explain.

Disjoint if … P (A∨B )=P ( A )+P (B)

0.7?¿

0.3+0.6

0.7≠0.9 Not disjoint! (There is an intersection)

P (A∨B )=P ( A )+P (B )−P ( A∧B )

0.7=0.3+0.6−P(A∧B)

0.7=0.9−P (A∧B)

−0.2=−P(A∧B)

0.2=P (A∧B) Intersection

7. Given two events A∧B where P (A )=0.1 , P (B )=0.7 ,∧P ( A∨B )=0.8, determine if the two events are disjoint. Explain.

Disjoint if … P (A∨B )=P ( A )+P (B)

0.8?¿

0.1+0.7

0.8=0.8 Disjoint! (There isn’t an intersection)

P (A∨B )=P ( A )+P (B )−P ( A∧B )

0.8=0.1+0.7−P(A∧B)

0.8=0.8−P (A∧B)

0=P(A∧B) So no Intersection

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Homework 15.4: Proving Independence and Disjoint Events

1. The table below shows the results of a survey in which students in 11th grade were asked if they participate in a sport. Determine if randomly selecting a male student and randomly selecting a student who participates in sports are independent events. Justify your answer.

2. The seniors from Harrison High School are required to participate in exactly one after-school sport. Data were gathered from a sample of 120 students regarding their choice of sport. The following data were recorded.

a. For this group of students, does this data suggest that gender and sports are independent of each other? Justify your answer.

b. Two students are chosen at random from 120 students. Find the probability that both play tennis.

3. 12% of U.S. Homes own a MAC computer and 72% of U.S. homes own at least two flat screen televisions. If the two events are independent, what is the probability of owning a MAC computer and owning at least two flat screen televisions?

(1) 9% (2) 60% (3) 17% (4) 94%24

4. Mrs. Reynolds wanted to see how the amount of homework her students completed was related to their test averages for the year. She organized the following data for her 110 students using an online grade book program. Use the following two-way frequency table to answer the following questions below:

Determine each of the following probabilities for a student chosen randomly from her group of students:

a. Average 70% - 85%:

b. Average 85% - 100% and less than 70% of homework complete:

c. 85% - 100% of homework complete or average 85% - 100%:

d. Average 70 or higher given that the student does 85% - 100% of the homework:

5. In the real estate ads, 64 % of homes have garages, 21% have swimming pools, and 17 % have both features.

a. What’s the probability a home has a garage or pool?

b. Are they mutually exclusive? Justify your answer.

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Lesson 15.5: Conditional Probability

Learning Goals:

1) What is conditional probability?2) How do you find conditional probability with/without two-way tables?

Warm Up: Answer the following questions in order to prepare for today’s lesson.

1. There is a red biscuit tin on a shelf. The tin contains three chocolate biscuits and seven plain biscuits. A child reaches into the red tin and randomly selects a biscuit. The child returns that biscuit to the tin, shakes the tin, and then selects another biscuit. Find the probability that both biscuits chosen are chocolate.

Independent! (child returns) 310

∗3

10= 9

100

2. There is a red biscuit tin on a shelf. The tin contains three chocolate biscuits and seven plain biscuits. A child reaches into the red tin and randomly selects a biscuit. The child eats that biscuit, shakes the tin, and then selects another biscuit. Find the probability that both biscuits chosen are chocolate.

Dependent! (child eats)310

∗2

9= 6

90= 1

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What are the differences about the two questions above? #1 the biscuit is returned and #2 the biscuit is eaten

How did this difference impact the probability of choosing two chocolate biscuits? In #1, the probability didn’t change and in #2 the probability changed

In question #1, the events of choosing two chocolate biscuits are considered independent events. What does this mean? Pick #1 didn’t impact pick #2

In question #2, the events of choosing two chocolate biscuits are considered dependent events? What does this mean? Pick #1 did impact pick #2

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Example: In a school of 1200 students, 250 are seniors, 150 students take math, and 40 students are seniors and also taking math.

a. What is the probability that a randomly chosen student is taking math, given they are a senior?

Total = 1200 P (M given S )=P(M∧S)P(S)

=

401200250

1200

= 40250

= 425

P (S )= 2501200

P (M )= 1501200

P (M∧S )= 401200 “given they are a senior” means it already occurred!

b. What is the probability that a randomly chosen student is a senior, given they are taking math?

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Definition

Conditional Probability- arise naturally in the investigation of experiments where an outcome of a trial may affect the outcomes of the subsequent trials. We try to calculate the probability of the second event (event B) given that the first event (event A) has already happened.

In words: The probability of an event (A), given that another event (B) has already occurred.

In symbols: P (A givenB )=P (A∧B)P(B) or P ( A|B )=P (A∧B)

P (B)

P (B given A )=P (B∧A)P (A ) or P (B|A )=P (B∧A)

P(A)

P (S givenM )=P(S∧M )P (M )

=

401200150

1200

= 40150

= 415

1. At a school, 60% of students buy a school lunch. Only 10% of students buy lunch and dessert. What is the probability that a student who buys lunch also buys dessert?

“student who buys a lunch” means it already occurred so is the “given” and makes it a conditional probability!

P (L )=.60 P (Dgiven L )=P(D∧L)P (L)

=0.100.60

=1060

=16

P (L∧D )=.10

2. A credit card company states that 425 of its customers are classified as long-term cardholders, 35% pay their bills in full each month, and 23% are long-term cardholders who also pay their bills in full each month. Let the event that a randomly selected customer is a long-term cardholder be L and the event that a randomly selected customer pays his bill in full each month be F.

a. What are the values of P (L ) , P (F ) ,∧P (L∧F)?

P (L )=0.42= 42100 P (F )=0.35= 35

100 P (L∧F )=0.23= 23100

b. Use the conditional probability formula to calculate P(L givenF). Round your answer to the nearest thousandth.)

P (Lgiven F )= P(L∧F)P (F )

=0.230.35

=0.657

c. Use the conditional probability formula to calculate P(F given L). (Round your answer to the nearest thousandth.)

P (F given L )= P(F∧L)P(L)

=

2310042

100

=2342

=0.548

Using a table to Find Conditional Probabilities: there are two ways to find conditional probability given a table.

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3. The table below shows the number of left and right handed tennis players in a sample of 50 males and females.

If a tennis player was selected at random from the group, find the probability that the player is:

a) female and right handed; 1650 b) left handed; 5

50

c) right handed, given that the player selected is female.

P (R givenF )= P(R∧F)P(F )

=

16501850

=1618

=89

d) right handed, given that the player selected is male.

P (R givenM )=P(R∧M )P (M)

=

29503250

=2932

4. A quality-control inspector checks for defective parts. The table shows the results of the inspector’s work. Find:

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a) the probability that a defective part “passes” already given!

P (P givenD )=P(P∧D)P (D)

=

350039

500

= 339

b) the probability that a non-defective part “fails”

[this also means the part will fail inspection, given that is is a non-defective part]

P (F givenND )=P (F∧ND )P(ND)

=

11500461500

= 11461

=.023

5. When a room is randomly selected in a downtown hotel, the probability that the room has a king-sized bed is 0.62, the probability that the room has a view of the town square is 0.43, and the probability that it has a king-sized bed and a view of the town square is 0.38. Let A be the event that the room has a king-sized bed, and let B be the event that the room has a view of the town square.

a) What is the meaning of P(A givenB) in this context?

b) Use a hypothetical 1000 table to calculate P(A givenB).

c) There is also a formula for calculating conditional probability. The formula for conditional probability is

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P (A givenB )=P (A∧B)P(B)

Use this formula to calculate P(A givenB), where the events A and B are as defined in this example.

At a restaurant, 25% of customers order chili. If 4% of customers order chili and a baked potato, find the probability that someone who orders chili also orders a baked potato. P(P givenC)

Park Medical Center currently has 200 patients. There are 40 patients diagnosed with lung cancer, 30 patients who are chronic smokers, and 25 patients who have lung cancer and smoke. If the patient smokes, what is the probability that the patient has lung cancer? P(LC given S)

6. Of the light bulbs available at a store, 42% are fluorescent, 23% are labeled as long life, and 12% are fluorescent and long life.

a. A light bulb will be selected at random from the light bulbs at this store. Rounding your answer to the nearest thousandth where necessary, find the probability that the selected light bulb is fluorescent given that it is labeled as long life. Conditional!

P (F )=0.42 P (F given L )= P(F∧L)P(L)

=0.120.23

=0.5217

P (L )=0.23

P (F∧L )=0.12

b. Are the events “fluorescent” and “long life” independent? Explain. They are independent if …P (F given L )=P(F )

0.5217≠0.42, so they are NOT independent

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Homework 15.5: Conditional Probability

1. When an avocado is selected at random from those delivered to a food store, the probability that it is ripe is 0.12, the probability that it is bruised is 0.054, and the probability that it is ripe and bruised is 0.019.

(a) Complete the hypothetical 1000 table given below, using the information given above.

(b) Rounding your answers to the nearest thousandth where necessary, find the probability that an avocado randomly selected from those delivered to the store is:

i. Not bruised.

ii. Ripe given that it is bruised.

iii. Bruised given that it is ripe.

(c) Are the events “ripe” and “bruised” independent? Explain your answer.

2. The probability that Gary and Jane have a child with blue eyes is 0.25, and the probability that they have a child with blond hair is 0.5. The probability that they have a child with both blue eyes and blond hair is 0.125. Determine the probability, to the nearest tenth of a percent, that a child has blond hair, given that they have blue eyes.

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3. Given events K∧L, such that P (K )=2.3 % ,P (L )=8.6 % ,∧P (K ∩L )=1.9%. Determine P(L|K )

4. At a gas station, 84% of customers buy gasoline. Only 5% of customers buy gasoline and a beverage. What is the probability that a customer who buys gasoline also buys a beverage?

5. The table below represents the participation in after school athletics by gender. Use this data to answer the following questions:

a. What is the probability that if a randomly selected student is female, she participates in the after school athletics program?

b. A randomly selected student is male. What is the probability he does not participate in the after school athletics program?

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Lesson 15.6: Proving Independence with Conditional Probability

Learning Goal: How do we use conditional probability to prove that two events are independent?

Warm-up: Answer the following questions in order to prepare for today’s lesson.

1. In New York State, 35% of the houses have a hot tub, 44 % of the houses have a pool, and 23 % of those houses have a pool and a hot tub. What is the probability that a house has a hot tub, given that it has a pool?

A=hot tub P (A )=.35 P (A given B )=P (A∩B)P(B)

= .23.44

= 2344

B=pool P (B )=.44

P (A∩B )=.23

2. The table below shows the results of a survey in which young adults ages 18−24 were asked if they ever used Instashop (Yes or No). Is the gender of young adults independent of their use of Instashop? Justify your answer.

A= Female P (A ) ∙ P (B )?¿P (A ∩B)

B= NOT uses 270510

∙ 122510

?¿

54510

.1266435986≠ .1058823529

So they are not independent (or dependent)!

3. Sean’s team has a baseball game tomorrow. He pitches 50 % of the games. There is a 40 % chance of rain during the game tomorrow and the probability that it rains given Sean pitches is 40 %. Using this data, determine whether the events “Sean pitches in the game” and “it rains during the game” are independent of each other? Justify your answer.

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P (A ) ∙ P (B )?¿ P (A ∩B) P (B given A )=P(B)

.50 ∙.40=? .40=.40

Can’t use this test!

4. In a class of students, the probability that a student is on the honor roll is 1140 and the

probability that the student is an athlete is 1240 . If the probability that a student is on the

honor roll given that they are an athlete is 1040 . It can be concluded that these two

events are

(1) independent (2) dependent (3) mutually exclusive (4) complements

P (A givenB )=P(A)

1040

≠ 1140 so dependent

5. Given: Events A∧B, such that P (A )=0.15 , P (B )=0.40 ,∧P ( A∩B )=0.06. Calculate P( A|B). Using your calculated probability, determine if events A∧B are independent or dependent.

P (A givenB )=P ( A∩B )P (B )

= .06.40

=.15 P (A givenB )=P (A )

.15= .15 so independent

6. The guidance department has reported that of the senior class, 2.3 % are members of key club, K , 8.6 % are enrolled in AP Physics, P, and 1.9 % are in both. Determine the probability of Pgiven K, to the nearest tenth of a percent. The principal would like to know if these two events are independent of each other. Using your calculated probability, write a statement relating student enrollment in the given situation.

P (K )=.023

P (P )=.086 P (P givenK )=P (P∩K )P (K )

= .019.023

=.8260869565=82.6 %

P (K∩P )=.019 35

Conditional Test for Independence (must memorize)

If two events are independent, then P (A givenB )=P (A )∨P (B given A )=P(B)

P (P givenK )=82.6 %

P (P givenK )=P(P)

.826≠ .086 therefore, Key Club and Physics are not independent (or dependent)!

Product Test for Independence

P (A ∩B )=P(A)× P(B)

Conditional Test for IndependenceP (A givenB )=P(A)

Disjoint Events (Mutually Exclusive)P (A∪B )=P ( A )+P (B)

7. A student chosen at random from the student body at a given high school. The

probability that the student selects Math as the favorite subject is 14 and the probability

that the student chosen is a junior is 116459 . The probability that the student selected is a

junior and chooses Math as the favorite subject is 31459 . Given this information, the two

events are I. dependent II. independent III. mutually exclusive

(1) I, only (2) II, only (3) I and III (4) II and III

8. Events A∧B have probabilities P (A )=0.4 ,P (B )=0.65 ,∧P (A∩B )=0.2.

a. Calculate P( A|B).

b. State with a reason whether events A∧B are independent.

c. State with a reason whether events A∧B are mutually exclusive.

Not mutually exclusive because P(A∩B)≠0

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Homework 15.6: Proving Independence with Conditional Probability

1. A credit card company states that for all of their customers, 42 % are classified as long-term cardholders, 35 % pay their bills in full each month, and 23 % do both.

Determine the probability, to the nearest tenth of a percent, that the customer is a long-term cardholder, given that they pay their bills in full each month.

Using your calculated probability, determine if the two events are independent.

2. Events A∧C have probabilities P (A )=0.15 , P (C )=0.40 ,∧P ¿. State with reason whether events A∧B are independent.

3. Sean’s team has a baseball game tomorrow. He pitches 50 % of the games. There is a 40 % chance of rain during the game tomorrow. If the probability that it rains and that Sean pitches is 20%, it can be concluded that these two events are

(1) independent (2) dependent (3) mutually exclusive (4) complements

4. Data collected about jogging from students with two older siblings are shown in the table below. Using these data, determine whether a student with two older siblings is more likely to jog if one sibling jogs or if both siblings jog. Justify your answer.

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Lesson 15.7: Venn Diagrams

Learning Goal: How can we represent two events with a Venn diagram? How can we use a Venn diagram to find the probability of two events?

Use the above diagram to help answer the questions below.

The Venn diagram below represents the students who participate in sports and band at a particular high school.

a. How many students participate in organized sports? 442+21=463

b. How many students play in the band? 21+31=52

c. How many students do not participate in organized sports? 31+339=370

d. How many students participate in both organized sports and band? 21

e. How many students participate in organized sports or play in the band?

442+21+31=494

Notation and Venn Diagrams38

Considering all the people in the world, let A be the set of Americans (citizens of the United States), and let B be the set of people who have brothers.

The set of people who are Americans and have brothers is represented by the shaded region in the Venn diagram below.

Shaded region is A∧B

⋂→ ∧”(¿)

The set is written as A⋂B and is read as A and B.

The set of people who are Americans or have brothers is represented by the shaded region in the Venn diagram below.

Shaded region is A∨B

⋃→ ∨”(¿)

The set is written as A⋃B and is read as A or B.

The set of people who are not Americans is represented by the shaded region in the Venn diagram below.

Not Americans is the complement or everything NOT in circle A

A'→NOT acomplement

The set is written as A ' and is read as A complement.

When Given the Intersection:

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1. A squash club has 27 members. 19 have black hair, 14 have brown eyes and 11 have both black hair and brown eyes.

a. Place this information on a Venn Diagram.

b. Find the number of members with black hair or brown eyes. 8+11+3=22

c. Find the number of members with black hair, but not brown eyes. 8

2. In a school of 320 students, 85 students are in band, 200 students are on sports teams, and 60 students participate in both activities.

a. Draw and label a Venn diagram to represent the situation.

b. How many students are involved in either band or sports?

40

3.: Suppose that 230 students play soccer, 190 students play basketball, and 60 students play both sports. There are a total of 500 students at the school.

a. Complete the Venn diagram below by writing the numbers of students in the various regions of the diagram.

b. Suppose that a student will be selected at random from the school.

i. What is the probability that the selected student plays both sports?

ii. Complete the Venn diagram below by writing the probabilities associated with the various regions of the diagram.

Using Probabilities in Venn Diagrams

Based on what you know about probabilities, what should the numbers in the Venn diagram add up to when working with probabilities? Sum ¿1∨100 %

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4. When a fish is selected at random from a tank, the probability that it has a green tail is 0.64, the probability that it has red fins is 0.25, and the probability that it has both a green tail and red fins is 0.19.

a. Draw a Venn diagram to represent this information.

b. Find the following probabilities:

i. The fish has red fins but does not have a green tail.

ii. the fish has a green tail but not red fins.

iii. The fish has neither a green tail nor red fins.

c. Complete the table below showing the probabilities of the events corresponding to the cells of the table.

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Homework 15.7: Venn Diagrams

1. In a company, 43% of the employees have access to a fax machine, 38% have access to a fax machine and a scanner, and 24% have access to neither a fax machine nor a scanner. Suppose that an employee will be selected at random. Using a Venn diagram, calculate the probability that the randomly selected employee will not have access to a scanner.

2. Kevin will soon be taking exams in math, physics, and French. He estimates the probabilities of his passing these exams to be as follows:

Math: 0.9 Physics: 0.8 French: 0.7

Kevin is willing to assume that the results of the three exams are independent of each other. Find the probability of each event.

a. Kevin will pass all three exams.

b. Kevin will pass math but fail the other two exams.

c. Kevin will pass exactly one of the three exams.

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3. In the real estate ads, 64% of homes have garages, 21% have swimming pools, and 17% have both features.

a. Represent this information on a Venn diagram.

b. What’s the probability a home has a garage or pool?

c. Are these two events disjoint? Justify your answer.

4. If E∧F are disjoint and P (E )=.4 ,P (F )=.5, compute P (E∧F )∧P(E∨F).

5. The table below describes the smoking habits of a group of asthma sufferers.

If one of the people is randomly selected, find the probability that the person is a man or a heavy smoker.

Are the events “women” and “occasional smoker” independent events? Justify your response.

6. Consider the Venn diagram below showing the probabilities of events A∧B.

a. Find each probability:

i. P (A )44

ii. P (B )

iii. P (A∨B )

b. Are events A∧B independent? Justify your answer.

c. Are events A∧B disjoint? Justify your answer.

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