1. Chapter 5 (Probability) Summary
David Cheuvront, Ruben Andrade, Noel Casino

2. What is Probability?
Probability is the chance that a certain outcome of a random process that describes how many times it will happen over the course of many trials.
We can use probability because of the large law of numbers, which states that the proportion of times that a particular outcome occurs in many repetitions will approach a single number.
Probability can be expressed in a few ways: ½, .50, and 50%. All these values are the same.
An example of probability is the chance of getting heads when you toss a fair coin. There is a .50 or 50% chance that you will get heads.
Remember that probability is a long-run phenomena. Regularity only emerges after many repetitions.

3. Probability Problem
Suppose the probability of bouncing a ping-pong ball into a red solo cup is 0.4.
What does this probability mean?
Answer: The probability in this situation means that over the course of many trials, the probability that the ping-pong ball will land in the cup is 0.4 or 40%.
Question B. Because the probability is 0.4, am I guaranteed to make 4 out of 10 shots?
Answer: No. The amount of times you make it into the cup could be greater or less than 4 because of random chance.

4. Basic Rules of Probability
The Probability of any event is a number between 0 and 1
All possible outcomes together must have probabilities that add up to 1.
If all outcomes in the sample space are equally likely, the probability that event A occurs can be found using the formula: Outcomes with Event A / Total Number of Outcomes.
The probability that an event does not occur is 1 minus the probability that the event does occur.
If two events have no outcomes in common, the probability that one or the other occurs is the sum of their individual probabilities.

5. Basic Rules of Probability Problem
Determine if the probability given to the following situations are legitimate, that is if it satisfies the rules of probability.
Pick a random person and record their gender and employment status, P(Woman, Employed) = 0.40, P(Woman, Employed) = 0.10, P(Man, Unemployed) = 0.05, P(Man, Employed) = 0.50.
Answer: They are not legitimate because = 1.05, which
does not add up to zero
Pick a random card from a deck. Record the suit the card is in.
P(Spade) = 13/52, P(Heart) = 12/52, P(Club) = 14/52, P(Diamond) = 13/52.
Answer: (13/52) + (12/52) + (14/52) + (13/52) = 52/52 = 1
This situation is legitimate because the probabilities all add up to 1.

6. Simulations
A simulation is an imitation of chance behavior. To perform one, follow these 4 steps:
State: Ask question about the chance process.
Plan: Describe how to use chance to imitate one repetition of the process. Decide what you want to record at the end of each repetition.
Do: Perform many trials of the simulation.
Conclude: Use results to answer the question of interest you created in step 1.

7. Simulation Problem
Suppose the chance of Ruben catching an iguana when he throws a net is 0.25.
How could we simulate his chance of catching an iguana with:
A deck of Cards
Answer: Let hearts represent him catching an iguana, and clubs and spades represent him not catching one. Draw a card.
Table D - line 112
Answer: Go to table D – line 112. Let numbers represent him catching an iguana, and represent him not catching one. Read a two-digit number from Table D line 112.
Using a random number generator.
Answer: Go to a random number generator and set the upper to 4 and the lower to 1. Let the number 1 represent that he catches an iguana and numbers 2-4 represent that he doesn’t.

8. The Probability Model for this problem is:
Probability Models
Suppose we flip a fair coin 3 times. Let X=The amount of times you get a head create the sample space and probability distribution for X.
Sample Space:
A sample space is the set of all possible outcomes.
The sample space for this problem is:
HHH TTT
HTH THT
HHT TTH
HTT THH
Probability Model:
The probability Model is a description of some chance process that consists of the sample space and the probability for each outcome.
The Probability Model for this problem is:
HHH TTT
HTH THT
HHT TTH
HTT THH
Each outcome has a probability of 1/8 of happening. The probability of all outcomes should add up to 1.

9. Probability Models Cont.
We can graph the probability of X = number of heads you get in 3 Coin Flips.
P(X=0) – 1/8
P(X=1) – 3/8
P(X=2) – 3/8
P(X=3) – 1/8
An event is any collection of outcomes from the same chance process. This is a subset of the sample space. Events are usually designated by capital letters, such as A,B,C, etc.
Define event A = “2 heads”
There are 3 outcomes in event A: HTH, HHT, THH
Therefore, the probability of Event A is 3/8 or

10. Mutual Exclusivity and the Compliment Rule
Two events are Mutually Exclusive, or disjoint, if they can never occur together. I.E. P(A and B) = 0
The Complement Rule is P(AC) = 1- p(A). Because we know that events of A occurring and not occurring always add up to 1, we know that the probability of A not happening is 1 – the probability of A occurring. We can use this rule to find out the complement of A, which is the probability of A not occurring.

11. Mutually Exclusive Problem
We sampled 400 truck owners at a chili cook-off sponsored by Chevy and Ford. Only trucks made by Chevy or Ford were let in to the event. In the following table, what event is mutually exclusive?
Answer: Having a truck made by Ford and having a 4 wheel drive is mutually exclusive because the probability of having a four wheel drive truck made by ford is 0. This could be a result of Ford not making a truck with a 4 wheel drive.
Truck Brand
4 Wheel Drive
2 Wheel Drive
Chevy
186 59
Ford
155

12. Complement Rule Problem
The probability of a randomly sampled American being obese is What is the probability of a randomly sampled American NOT being obese?
Answer: By using the complement rule, we can find the answer to this question.
We know that the complement rule is P(AC) = 1- p(A). To answer this problem, we need to find P(AC).
P(AC) = 1 – 0.40
P(AC) = 0.60
The answer to this question is that the probability of a randomly sampled American NOT being obese is 0.60.

13. Two-Way Tables
Two-Way tables help display the probability two events happening.
In this two way table, we display the number of people that are both married and their genders. This helps display two variables and their outcomes easily.
Married
Male
Female
Total
Yes
218
316
534 No
180 86
266
398
402
800

14. Completing a Two-Way Table Problem
Complete the Two-way Table shown below.
Answer: To complete the two way table, we must find the amount of women living in a house. To do this, we subtract the total number of people living in a house and the number of men living in a house.
249 + ? = 448
448 – 249 = 199
The number of females living in a house is 199. Using the information given to us in this incomplete table, we can complete it.
Type of Housing
Male
Female
Total
House
249
???
448
Apartment 92
160
252
341
359
700

15. “And” Notation
If we want to find the probability of something having one characteristic AND another characteristic, we find the probability of (A and B). The notation for this is P(A Ω B). This means that the outcome has to meet both the requirements of A and B.
We will use the two-way table from question 6 to demonstrate what we are looking for. Say we want to find the probability of a person that is married AND a male, or P(Married Ω Male). We would try to find the probability of randomly selecting the highlighted cell.
Married
Male
Female
Total
Yes
218
316
534 No
180 86
266
398
402
800

16. “And” Notation Continued
To do this, we would put the number of people that fit the requirements of being married AND being a male, and divide it by the total amount of people.
218/800 =
P(Married Ω Male) = or 27.25%

17. “And” Notation Problem
Use the table below to find P(Dog Ω Male).
Answer: To find this probability, we have to find the number of animals that are both a dog AND a male and divide it by the total number of animals.
P(Dog Ω Male) = 572/1000
P(Dog Ω Male) = or 57.2%
Animal
Male
Female
Total
Dog
572
228
800
Cat 97
103
200
669
331
1000

18. “Or” Notation
If we want to find the probability of something of having one characteristic OR another characteristic, we find the probability of (A or B). The notation for this is P(A U B). We use this whenever we want to find out when an outcome meets the requirements of A or B.
We will use the two-way table from question 6 again to demonstrate what we are looking for. Say we want to find the probability of a person that is married OR a male, or P(Married U Male). We would try to find the probability of randomly selecting the highlighted cells.
Married
Male
Female
Total
Yes
218
316
534 No
180 86
266
398
402
800

19. “Or” Notation Continued
To find the probability, we would add up the number of people that fit the requirement of being married OR being male and then divide it by the total number of people
( )/800 =
P(Married U Male) = or 89.25%
*Be careful not to include the cell highlighted in purple twice! If you did, your probability would be over 1, which, as previously stated, is impossible!*

20. “Or” Notation Problem Use the following table to find P(Deer U Male).
Answer: To find the probability, we must add up the number of all the animals that are either a deer OR a male and divide it by the total number of animals.
P(Deer U Male) = ( ) / 1000
P(Deer U Male) = or 81.4%
*Remember not to count the same value twice in your calculations!)*
Animal
Male
Female
Total
Deer
324
348
672
Moose
142
186
328
466
534
1000

21. Venn Diagrams
Venn Diagrams are very similar to two-way tables, as they show the probability of two variables.

22. Conditional Probability
Conditional Probability is the probability that one event happens given that another event is already known to have happened. Lets say that we know that event B already happened. The probability that event B happens given that event A has happened is denoted by P(A І B)
To find the conditional probability of an event, use the formula:
P(A І B) = P(A Ω B) / P(B)

23. Conditional Probability Problem
Use the following table to find P(Male І Home).
Answer: We are trying to find the probability of a person being a male, given that they have a home. To do this, we must find the P(Male Ω Home) and divide it by P(Home).
P(Male І Home) = (44/100) / (89/100)
P(Male І Home) = or 49.44%
Home?
Male
Female
Total
Home 44 45 89
Homeless 8 3 11 52 48
100

24. The General Multiplication Rule
The General Multiplication Rule is used to find the probability that events A and B both occur.
The formula for the Rule is: P(A Ω B) = P(A) x P(A І B)
This rule is useful when a chance process involves a sequence of outcomes. In these cases, we can use tree diagrams to display the sample space.

25. General Multiplication Rule Problem
According to our research, the probability that Noel is from Vietnam is The probability of a randomly selected person from Vietnam being Chinese is What is the chance that Noel is from Vietnam and Chinese?
We know that P(Vietnam) = 0.64 and P(Vietnam І Chinese) = 0.24
According to our general multiplication rule, P(Vietnam Ω Chinese) = P(Vietnam) x P(Vietnam І Chinese).
P(Vietnam Ω Chinese) = 0.64 x 0.24
P(Vietnam Ω Chinese) = or 15.36%

26. Tree Diagrams
Tree diagrams rely heavily on the multiplication rule. They show all the probabilities of each route.

27. Independent Events
Two events are independent if the occurrence of one event does not change the probability that the other event will happen. Events A and B are independent if P(A І B) = P(A) x P(B І A) = P(B)
The multiplication rule for independent events is:
P(A Ω B) = P(A) x P(B)

28. Question 1!
P(X) = 0.25 and P(Y) = If P(X I Y) = 0.20, what is P(Y I X)?
0.10
0.125
0.32
0.45
0.50

29. Question 1 Answer The answer is C: 0.32
P(X Ω Y) = P(X I Y) x P(Y) = 0.20 x 0.40 = 0.08
Then P(Y I X) = P(X Ω Y) / P(X) = 0.08 / 0.25 = 0.32

30. Question 2!
Given the probabilities P(A) = 0.3 and P(A U B) = 0.7, what is the probability P(B) if A and B are mutually exclusive? If A and B are independent?
0.4, 0.3
0.4, 4/7
4/7, 0.4
0.7, 4/7
0.7, 0.3

31. Question 2 Answer The answer is B: 0.4, 4/7
If A and B are mutually exclusive, P(A Ω B) = 0. Thus, 0.7 = P(B) – 0, and so P(B) = 0.4.
If A and B are independent, then P(A Ω B) = P(A) x P(B). Thus 0.7 = P(B) -0.3P(B), and so P(B) = 4/7

32. Question 3
You win a game if you flip a coin with heads coming up exactly 50% of the tosses, Would you rather flip 10 times or 100 times?
10 times because > 0.080
10 times because of the Central Limit Theorem
100 times because of the Law of Large Numbers.
Your chance is the same.

33. Question 3 Answer The answer is A: 10 times because 0.246 > 0.080
Binomcdf (10, 0.5, 5) = while Binomcdf(100, 0.5, 50) = The probability that you get exactly 50% of the coin tosses being heads on throwing just 10 coins is higher than if you were to do it 100 times.

34. Question 4!
Given two events, E and F, such that p(E) = 0.340, P(F) = 0.450, and P(E U F) = 0.637, then the two events are:
Independent and mutually exclusive.
Independent, but not mutually exclusive.
Mutually Exclusive, but not independent.
Neither independent nor mutually exclusive.
There is not enough information to determine the answer.

35. Question 4 Answer
The answer is B:Independent, but not mutually exclusive.
P(E U F) = P(E) = P(F) – P(E Ω F), so = – P(E Ω F) and P(E Ω F) = Since P(E Ω F) does not equal 0, E and F are not mutually exclusive.
P(E Ω F) = = x = P(E) x P(F), which implies that E and F are independent.

36. Question 5!
There are two games involving flipping a fair coin. In the first game, you win a prize if you can throw between 45% and 55% heads; in the second game, you win if you can throw more than 60% heads. For each game, would you rather flip the coin 30 times or 300 times?
30 times for each game
300 times for each game
300 times for the first game, 30 times for the second game
The outcome of the games do not depend on the number of flips.

37. Question 5 Answer
The Answer is C: 300 times for the first game, 30 times for the second game
By the Law of Large Numbers, the more times you flip a coin, the closer the relative frequency tends to be to the probability of With fewer tosses, there is more chance of wide swings in the relative frequency.

38. Question 6!
A basketball player makes one out of his first two free throws. From that point on, the probability that he makes the next shot is equal to the proportion of shots made up to that point. If he takes two more shots, what is the probability he ends up making a total of two free throws?
1/4
1/3
1/2
2/3
3/4

39. Question 6 Answer The Answer is B: 1/3
P(2 shots made) = (1/2) x (1/3) + (1/2) x (1/3)