Chapter 6:

Probability and

Simulation The study of randomness

Introduction

Probability is the study of chance.

6.1 focuses on simulation since actual observations are often not feasible.

When we produce data by random sampling or randomized comparative experiments laws of probability answer the question “what would happen if we did this many times?”

Probability is the basis of inference

6.1 Simulation

The experiment consists of spinning the spinner

three times and recording the numbers as they

occur (e.g. 123). We want to determine the

proportion of times that at least one digit

occurs in its correct position. (we will use a

calculator instead).

Guess the proportion of times at least one digit

will occur in its proper place.

Get your calculators out!!

Enter the command “randint(1,3,3)”

Continue to press “enter” to generate

more three-digit numbers

Record the result in a table

At least one digit in the correct position

None of the digits in the correct position

The more repetitions, the closer a result’s

occurrence will get to it’s true likelihood.

Independence: When the result of one

trial (coin toss, dice roll) has no effect or

influence on the next toss.

Coin Toss

Simulation Steps 1. State the problem or describe the random

phenomenon Ex: Toss a coin 10 times, what is the likelihood of a run of at least 3

consecutive heads or 3 consecutive tails?

2. State the Assumptions (there are 2) A head or a tail is equally likely to occur on each toss

Tosses are independent of each other

3. Assign digits to represent outcomes (want efficiency)

In a random # table, even and odd digits occur with the same

long-term relative frequency (50%)

One digit simulates one toss of the coin

Odd digits represent heads; even digits represent tails

Successive digits in the table simulate independent tosses

4. Simulate many repetitions Looking at 10 consecutive digits in Table B simulates one repetition.

Read many groups of 1- digits from the table to simulate many rep

Be sure to keep track of whether or not the event we want (a run of at least 3 heads or at least 3 tails) occurs on each repetitions Here are the first 3 repetitions starting at line 101 in Table B.

Digits: 19223 95034 05756 28713 96409 12531

H/T: HHTTH HHTHT THHHT TTHHH HTTTH HTHHH

Run of 3: YES YES YES

22 repetitions were done for a total of 25. 23 of them did have a run of 3 or more Heads or tails.

5. State your conclusions We estimate the probability of a run of size 3 by the proportion

23/25 =.92

*Of course 25 reps are not enough to be confident that our estimate is accurate so we can tell a computer to do thousands of repetitions (or TRIALS) for us. A long simulation finds that the true probability is .86

Assigning digits Some ways more efficient than others.

Example: Choose a person at random from a group of which 70% are employed. One digit simulates one person: 0, 1, 2, 3, 4, 5, 6 = employed

7, 8, 9 = not employed

00- 69 employed and 70-99 not employed could also have worked, but is less efficient b/c requires twice as many digits and ten times as many numbers.

Example 2: Choose one person at random from a group of which 73% employed Now 00-72 = employed, 73-99 = not employed

Example 3: Choose one person at random from a group of which 50% are employed, 20% are unemployed, and 30% are not in the labor force: 0-4 = employed, 5-6 = unemployed, 7-9 = not in labor force.

Frozen Yogurt Sales example

Orders of frozen yogurt flavors (based on

sales) have the following relative

frequencies: 38% chocolate, 42% vanilla,

20% strawberry.

We want to simulate customers entering

the store and ordering yogurt.

How would you simulate 1- frozen yogurt sales

based on recent history using table?

Randomizing with Calculator

Block of 5 random digits from table

Rolling a die 7 times

10 numbers from 00-99

0

1

0.01 0.6

You read in a book on poker that the

probability of being dealt three of a kind

in a five-card poker hand is 1/50. Explain

in simple language what this means.

If the hands were dealt many times,

about 2% (1out of 50) hands will contain a

three of a kind.

6.2 Probability Models

Proportion of heads to tails in a few tosses

will be erratic but after thousands of tosses

will approach the expected .5 probability

Probability models have two parts:

A list of possible outcomes

A probability for each outcome.

Sample Space

To specify S we must state what

constitutes an individual outcome, then

which outcomes can occur (can be

simple or complex)

Ex: coin tossing, S = {H, T}

Ex: US Census: If we draw a random sample

of 50,000 US households, as the survey does,

the S contains all 50,000

Rolling two dice At a casino- 36 possible outcomes when

we roll 2 dice and record the up-faces in

order (first die, second die)

Gamblers care only about number of dots

face up so the sample space for that is:

S = {2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12}

Techniques for

finding outcomes

1. Tree diagram

For tossing a coin

then rolling a die

2. Multiplication Principle

2x6 = 12 for same example

3. Organized list:

H1, H2, H3, H4, H5, H6, T1, T2, T3, T4, T5, T6

With/without replacement

If you take a card from a deck of 52,

don’t put it back, then draw your 2nd card

etc., that’s without replacement.

Ex: how many different 3 digit numbers can

you make: 10x9x8 = 720

If you take a card, write it down, put it

back, draw 2nd card etc., that’s with

replacement.

Ex: 10x10x10 = 1000

Probability Rules 1. Any probability is a number between 0 and 1

2. The sum of the probabilities of all possible outcomes = 1

3. If 2 events have no outcomes in common (they can’t occur together), the probability that one OR the other occurs is the sum of their individual probabilities Ex: If one event occurs in 40% of all trials, another event

happens in 25% of all trials, the 2 can never occur together, then one or the other occurs on 65% of all trials

4. The probability that an event doesn’t occur is 1 minus the probability that it does occur Ex: If an event happens in 70% of all trials, it fails to occur

in the other 30%

Venn diagrams help!

Ex: Probability of rolling a 5?

B/c P(roll a 5 with 2 die) =

P(1,4) + P(3,2) + P(2,3) + P(4,1)

= 1/36 + 1/36 + 1/36 + 1/36

= 1/9 or .111

Independence &

the Multiplication Rule To find the probability for BOTH events A and B

occurring

Example: Suppose you plan to toss a coin twice, and

want to find the probability of rolling a head on both

tosses.

A = first toss is a head, B = second toss is a head. So

(1/2)(1/2) = ¼. We expect to flip 2 heads on 25% of

all trials. The more times we repeat this, the closer our

average probability will get to 25%.

The multiplication rule applies only to independent

events; can’t use it if events are not independent!

Independent or not? Coin toss

I: Coin has no memory and coin tossers cannot influence fall of coin

Drawing from deck of cards

NI: First pick, probability of red is 26/52 or .5.

Once we see the first card is red, the probability

of a red card in the 2nd pick is now 25/51 = .49

Taking an IQ test twice in succession

NI

More applications of

Probability Rules

If two events A and B are independent,

then their complements are also

independent.

Ex: 75% of voters in a district are

Republicans. If an interviewer chooses 2

voters at random, the probability that the

first is a Republican and the 2nd is not a

republican is .75 x .25 = .1875

6.3 General Probability Rules

Addition Rule for Disjoint events

General Addition rule for

Unions of 2 events

Example:

Deb and Matt are waiting anxiously to hear if they’ve been promoted. Deb guesses her probability of getting promoted is .7 and Matt’s is .5, and both of them being promoted is .3. The probability that at least one is promoted = .7 + .5 - .3 which is .9. The probability neither is promoted is .1.

The simultaneous occurrence of 2 events (called a joint event, such as deb and matt getting promoted) is called a joint probability.

Conditional Probability The probability that we assign to an event

can change if we know some other event has occurred.

P(A|B): Probability that event A will happen under the condition that event B has occurred.

Ex: Probability of drawing an ace is 4/52 or 1/13. If your are dealt 4 cards and one of them is an ace, probability of getting an ace on the 5th card dealt is 3/48 or 1/16 (conditional probability- getting an Ace given that one was dealt in the first 4).

In words, this says that for both of 2 events to occur,

first one must occur, and then, given that the first

event has occurred, the second must occur.

Remember: B is the event whose probability

we are computing and A represents the info

we are given.

Extended Multiplication rules

The union of a collection of events is the

event that ANY of them occur

The Intersection of any collection of

events is the event that ALL of them occur

Example Only 5% of male high school basketball, baseball, and football

players go on to play at the college level. Of these only 1.7% enter major league professional sports. About 40% of the athletes who compete in college and then reach the pros have a career of more than 3 years. Define these events: A = competes in college B = competes pro C = pro career longer than 3

years

P(A) = .05

P(B|A) = .017

P(C|A and B) = .400

What is the probability a HS athlete will have a pro career more than 3 years? The probability we want is therefore P(A and B and C) = P(A)P(B|A)P(C|A and B)

= .05 x .017 x .40 = .00034

So, only 3 of every 10,000 high school athletes can expect to compete in college and have a pro career of more than 3 years.

Extended tree diagram + chat

room example 47% of 18 to 29 age chat online, 21% of 30 to 49

and 7% of 50+

Also, need to know that 29% of all internet users are 18-29 (event A1), 47% are 30 to 49 (A2) and the remaining 24% are 50 and over (A3). What is the probability that a randomly chosen

user of the internet participates in chat rooms (event C)?

Tree diagram- probability written on each segment is the conditional probability of an internet user following that segment, given that he or she has reached the node from which it branches.

(final outcome is adding all the chatting probabilities which = .2518)

Bayes Rule Another question we might ask- what percent of

adult chat room participants are age 18 to 29?

P(A1|C) = P(A1 and C) / P(C)

= .1363/.2518 = .5413

*since 29% of internet users are 18-29, knowing that someone chats increases the probability that they are young!

Formula sans tree diagram:

P(C) = P(A1)P(C|A1) + P(A2)P(C|A2) + P(A3)P(C|A3)

6.3 Need to Know summary(print) Complement of an event A contains all outcomes not in A

Union (A U B) of events A and B = all outcomes in A, in B, or in both A and B

Intersection(A^B) contains all outcomes that are in both A and B, but not in A alone or B alone.

General Addition Rule: P(AUB) = P(A) + P(B) – P(A^B)

Multiplication Rule: P(A^B) = P(A)P(B|A)

Conditional Probability P(B|A) of an event B, given that event A has occurred: P(B|A) = P(A^B)/P(A) when P(A) > 0

If A and B are disjoint (mutually exclusive) then P(A^B) = 0 and P(AUB) = P(A) + P(B)

A and B are independent when P(B|A) = P(B)

Venn diagram or tree diagrams useful for organization.