Chapter 6:

Probability and

SimulationThe study of randomness

Where we left off in 6.2…

Example of disjoint/complement Select a woman aged 25 – 29 years old at random

and record her marital status. At random means that we give every such woman the same chace to be the one we choose. We choose an SRS of size 1. The probability of any marital status is just the proportion of all women aged 25 to 29 who have that status; if we selected many women, this is the proportion we would get.

Here is the probability model

Find the probability that the woman we draw is not married, using The complement rule

The addition rule

Marital Status Never married Married Widowed Divorced

Probability 0.353 0.574 0.002 0.071

Probabilities in a finite space

Looking at the probability model re:

marital status of women, notice the sum

of the separate events.

The probabilities for the events ended up

being unique numbers, but if two events

have the same probability, they are

labeled as equally likely.

Independence &

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

occurring

If finding the probability that two events occur, the

probabilities of these events are multiplied.

The multiplication rule applies only to independent

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

In a Venn diagram, the event {A and B}is represented in the overlap

Independent or not? Examples 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

Multiplication Rule Example 1 A general can plan a campaign to fight

one major battle or three small battles.

He believes that he has probability 0.6 of

winning the large battle and probability

0.8 of winning each of the small battles.

Victories or defeats in the small battles are

independent. The general must win either

the large battle or all three small battles to

win the campaign. Which strategy should

he choose?

One big battle is a greater probability of

0.6 vs three small battles at 0.512

(0.8*0.8*0.8)

Multiplication Rule Example 2 A diagnostic test for the presence of the AIDS virus

has the probability of 0.005 of producing a false

positive. That is, when a person free of the AIDS virus is tested, the test has probability 0.005 of

falsely indicating that the virus is present. If all 140

employees of a medial clinic are tested and all

140 are free of AIDS, what is the probability that at

least one false positive will occur?

P(at least one positive)=1-P(no positives)

1-P(140 negatives)

1-(0.995^140)

1-0.496

0.504

6.3 General Probability Rules

Addition Rule for Disjoint eventsThe addition rule for disjointed events is used when finding the

probability of one event occurring. Event A OR B. Another word

frequently used for disjoint is mutually exclusive.

To be mutually

exclusive, two events

have no outcomes in

common; they can

never occur together.

EXAMPLE 1: Addition Rule

for Disjoint events

Randomly select a student who took the

2010 AP Statistics exam and record the

student’s score.

Find the probability that the chosen

student scored 3 or better.

Answer: 0.584

Score 1 2 3 4 5

Probability 0.233 0.183 0.235 0.224 0.125

EXAMPLE 2: General Addition

Rule for the Union of two events A random sample of people who participated in the 2000

Census was chosen. Suppose a member of the sample is

chosen at random. Find the probability that the member

a. …is a high school graduate

b. …is a high school graduate or owns a home

High School

Grad

Not a high

school grad

Total

Homeowner 221 119 340

Not a

homeowner

89 71 160

Total 310 190 500

Answer

a. 310/500 graduated from high school = 0.62

b. For b, it’s more complicated than just

adding…because the events are not mutually

exclusive

General Addition rule for

Unions of 2 events

b. In the case of b, the events of “high

school graduate” (event A) and “owning a

home” (event B) are non-mutually

exclusive, meaning that these events can

have outcomes in common (you can have

events A and B…you can be a homeowner

and a graduate of high school).

Seeing the overlap

High School

Grad

Not a high

school grad

Total

Homeowner 221 119 340

Not a

homeowner

89 71 160

Total 310 190 500

P(homeowner or hs grad) = P(homeowner)+ P(hs grad) – P(homeowner and hs grad)

= 340/500 + 310/500 – 221/500

= 429/500

Example 3:

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 or non-mutually exclusive, (such as deb and matt getting promoted)

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.

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.

Example 4: Conditional Probability

You choose a woman at random.

Find the probabilities:

a. P(married):

b. P(age 18 to 24 and married) :

c. P(married l age 18 to 24): the

bar is rad as “given the

information that…”Conditional Probability Solutions

You choose a woman at random. Find the probabilities:

a. P(married): 0.592

b. P(age 18 to 24 and married) : 0.031

c. P(married l age 18 to 24): 0.241

Remember: B is the event whose probability

we are computing and A represents the info

we are given.

Conditional Probability: the probability that one event happens given

that another event is already known to have happened. Suppose we

know that event A has happened. Then the probability that event B

happens given that event A has happened is denoted by P(B l A)

“The General Multiplication Rule for Two Events” can be rewritten if

we’re looking at finding the conditional probability, P(B l A).

Conditional Probability

Example 5 (use table)

What is the conditional probability that a

woman is a widow, given that she is at

least 65 years old?

Conditional Probability with

Multiplication Rules: Example 6

Slim is a professional poker player. At the

moment, he wants to draw two diamonds

in a row. At the table before him

between his cards and the cards of the

other players, there are 4 diamonds and 7

cards of other suits.

What is the probability that the next two

cards he draws are diamonds?

9/41 x 8/40 = 0.044

Example 7 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.

Tree Diagram Example 8

Make a tree diagram with corresponding

probabilities for the high school athletes

example, and use the different paths to

identify the probability that a high school

athlete will go on to professional sports.

Example 9: Extended tree diagram

+ chat room example 47% of 18 to 29 chat online, 21% are 30 to 49 years

old and 7% are 50+ years old

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)

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.