# Chapter 6 Probability and Simulation

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Chapter 6 Probability and Simulation

Simulation The imitation of chance behavior based on a model that accurately reflects the experiment under consideration, is called a simulation

Steps for Conducting a Simulation
State the problem or describe the experiment State the assumptions Assign digits to represent outcomes Simulate many repetitions State your conclusions

Step 1: State the problem or describe the experiment
Toss a coin 10 times. What is the likelihood of a run of at least 3 consecutive heads or 3 consecutive tails?

Step 2: State the Assumptions
There are Two A head or tail is equally likely to occur on each toss Tosses are independent of each other (ie: what happens on one toss will not influence the next toss).

Step 3 Assign Digits to represent outcomes
Since each outcome is just as likely as the other, and there you are just as likely to get an even number as an odd number in a random number table or using a random number generator, assign heads odds and tails evens.

Step 4 Simulate many repetitions
Looking at 10 consecutive digits in Table B (or generating 10 random numbers) simulates one repetition. Read many groups of 10 digits from the table to simulate many repetitions. Keep track of whether or not the event we want ( a run of 3 heads or 3 tails) occurs on each repetition. Example 6.3 on page 394

Step 5 State your conclusions. We estimate the probability of a run by the proportion Starting with line 101 of Table B and doing 25 repetitions; 23 of them did have a run of 3 or more heads or tails. Therefore estimate probability = If we wrote a computer simulation program and ran many thousands of repetitions you would find that the true probability is about .826

Various Simulation Scenarios
Example 6.4 – page Choose one person at random from a group of 70% employed. Simulate using random number table.

Frozen Yogurt Sales Example 6.5 – page 396 – Using random number table simulate the flavor choice of 10 customers entering shop given historic sales of 38% chocolate, 42% vanilla, 20% strawberry.

A Girl or Four Example 6.6 – Page 396 – Use Random number table to simulate a couple have children until 1 is a girl or have four children. Perform 14 Simulation

Simulation with Calculator
Activity 6B – page 399 – Simulate the random firing of 10 Salespeople where 24% of the sales force are age 55 or above. (20 repetitions)

Homework Read 6.1, 6.2 Complete Problems 1-4, 8, 9, 12

Chapter 6 Probability and Simulation
6.2 Probability Models

Key Term Probability is the branch of mathematics that describes the pattern of chance outcomes (ie: roll of dice, flip of coin, gender of baby, spin of roulette wheel)

Key Concept “Random” in statistics is not a synonym of “haphazard” but a description of a kind of order that emerges only in the long run

In the long run, the proportion of heads approaches
In the long run, the proportion of heads approaches .5, the probability of a head

Researchers with Time on their Hands
French Naturalist Count Buffon (1707 – 1788) tossed a coin 4040 time. Results: 2048 head or a proportion of English Statistictian Karl Person 24,000 times. Results 12, 012, a proportion of Austrailian mathematician and WWII POW John Kerrich tossed a coin 10,000 times. Results 5067 heads, proportion of heads .5067

Key Term / Concept We call a phenomenon random if individual outcomes are uncertain but there is nonetheless a regular distribution of outcomes in a large number of repetitions

Key Term / Concept The probability of any outcome of a random phenomenon is the proportion of times the outcome would occur in a very long series of repetition.

Key Term / Concept As you explore randomness, remember
You must have a long series of independent trials. (The outcome of one trial must not influence the outcome of any other trial) We can estimate a real-world probability only by observing many trials. Computer Simulations are very useful because we need long runs of data.

Key Term / Concept The sample space S of a random phenomenon is the set of all possible outcomes. Example: The sample space for a toss of a coin. S = {heads, tails}

The 36 Possible Outcomes in rolling two dice.

A Tree Diagram can help you understand all the possible outcomes in a Sample Space of Flipping a coing and rolling one die.

Key Concept Multiplication Principle - If you can do one task in n1 number of ways and a second task in n2 number of ways, then both tasks can be done in n1 x n2 number of ways. ie: flipping a coin and rolling a die, 2 x 6 = 12 different possible outcomes

Key Term / Concept With Replacement – Draw a ball out of bag. Observe the ball. Then return ball to bag. Without Replacement – Draw a ball out of bag. Observe the ball. The ball is not returned to bag.

Key Term / Concept With Replacement – Three Digit number
10 x 10 x 10 = 1000 ie: lottery select 1 ball from each of 3 different containers of 10 balls Without Replacement – Three Digit number 10 x 9 x 8 = 720 ie: lottery select 3 balls from one container of 10 balls.

Key Concept / Term An event is an outcome or a set of outcomes of a random phenomenon. An event is a subset of the sample space. Example: a coin is tossed 4 times. Then “exactly 2 heads” is an event. S = {HHHH, HHHT,………..,TTTH, TTTT} A = {HHTT, HTHT, HTTH, THHT, THTH, TTHH}

Key Definitions Sometimes we use set notation to describe events.
Union: A U B meaning A or B Intersect: A ∩ B meaning A and B Empty Event: Ø meaning the event has no outcomes in it. If two events are disjoint (mutually exclusive), we can write A ∩ B = Ø

Venn diagram showing disjoint Events A and B

Venn diagram showing the complement Ac of an event A

Complement Example Example 6.13 on page 419

Probabilities in a Finite Sample Space
Assign a Probability to each individual outcome. The probabilities must be numbers between 0 and 1 and must have a sum 1. The probability of any event is the sum of the outcomes making up the event Example 6.14 page 420

Assigning Probabilities: equally likely outcomes
If a random phenomenon has k possible outcomes, all equally likely, then each individual outcome has probability 1/k. The probability of any event A is P(A) = count of outcomes in A count of outcomes in S Example: Dice, random digits…etc

The Multiplication Rule for Independent Events
Rule 3. Two events A and B are independent if knowing that one occurs does not change the probability that the other occurs. If A and B are independent. P(A and B) = P(A)P(B) Examples: page 426

Homework Read Section 6.3 Exercises 22, 24, 28, 29, 32-33, 36, 38, 44

Probability And Simulation: The Study of Randomness
6.3 General Probability Rules

Rules of Probability Recap
Rule 1. 0 < P(A) < 1 for any event A Rule 2. P(S) = 1 Rule 3. Addition rule: If A and B are disjoint events, then P(A or B) = P(A) + P(B) Rule 4. Complement rule: For any event A, P(Ac) = 1 – P(A) Rule 5. Multiplication rule: If A and B are independent events, then P(A and B) = P(A)P(B)

Key Term The union of any collection of events is the event that at least one of the collection occurs.

The addition rule for disjoint events: P(A or B or C) = P(A) + P(B) + P(C) when A, B, and C are disjoint (no two events have outcomes in common)

General Rule for Unions of Two Events, P(A or B) = P(A) + P(B) – P(A and B)

Example 6.23, page 438

Conditional Probability
Example 6.25, page 442, 443

General Multiplication Rule
The joint probability that both of two events A and B happen together can be found by P(A and B) = P(A)P(B | A) P(A ∩ B) = P(A)P(B | A) Example: 6.26, page 444

Definition of Conditional Probability
When P(A) > 0, the conditional probability of B given A is P(B | A) = P(A and B) P(A) Example 6.28, page 445

Key Concept: Extended Multiplication Rule
The intersection of any collection of events is the even that all of the events occur. Example: P(A and B and C) = P(A)P(B | A)P(C | A and B)

Example 6.29, page 448: Extended Multiplication Rule

Tree Diagrams Revisted
Example 6.30, Page 448-9, Online Chatrooms

Bayes’s Rule Example 6.31, page 450, Chat Room Participants

Independence Again Two events A and B that both have positive probability are independent if P(B | A ) = P(B)

Homework Exercises #71-78, 82, 86-88