Assignment: Basic Probability Questions

p. 416

6.29,  6.33,  6.36

Add these to your last in class practice assignment.

Set Notation• A U B read “ A union B” is the set of all outcomes that are

either in A or in B. It is the same as A or B.

• The symbol Ø is used for the empty event, that is the event has no outcomes in it.

• read “A intersection B” is the set of all outcomes that are in A and B.

Venn diagram showing disjoint (mutually exclusive) events A and B.

Venn Diagram showing the complement Ac of an event A

Distance Learning Complement Rule

• Choose at random an undergraduate student taking a distance learning course for credit and record the student’s age. The probability of any age group is just the proportion of all distance learners in that age group--If we drew many students, this is the proportion we would get. Here is the probability model.

Age group

(year)18­23 24­29 30­39 40 or over

Probability 0.57 0.17 0.14 0.12

Complement rule cont…• The probability that the student we draw is not in the

traditional undergraduate age range is, by the complement rule,

P(not 18-23) = 1-P(18-23)

= 1-0.57=0.43

That is, if 57% of distance learners are 18-23 years old, then the remaining 43% are not 18-23.

Addition Rule• The events “30-39 years old” and “40 years or older” are

disjoint because no one can be in both groups.

• Therefore, the addition rule says

• P(30 years or older)= P(30-39) + P (40 years or older) =

• 0.14 + 0.12 = 0.26

Probability for Rolling a Dice

• For casino dice it is reasonable to assign equal probability (1/36) for each of the 36 outcomes.

• Gamblers are often interested in the total # of pips rolled.

• P(roll a 5) =1/36 + 1/36 +1/36 +1/36

• =4/36 =0.11

What is the probability of rolling a 7?

There are six outcomes where the sum is 7, therefore the probability is 6/36 =0.167.

Assigning Probabilities:

Finite # of Outcomes

Benford’s law• Faked numbers in a tax return, payment records, invoices

etc… often display patterns that are not present in legitimate records.

• The first digits of numbers in legitimate records often follow a distribution known as Benford’s law.

Investigators can detect fraud by comparing these probabilities with the first digits in records such as invoices paid by a business.

First digit 1 2 3 4 5 6 7 8 9

probability 0.301 0.176 0.125 0.097 0.079 0.067 0.058 0.051 0.046

A) Consider these events• A= (first digit is 1)

• B= (first digit is 6 or greater)

• From the table of probabilities, P(A)=P(1)=0.301

• P(B)= P(6)+P(7)+P(8)+P(9)

• =0.067+0.058+0.051+0.046=0.222

B) The probability that a first digit is something other than 1 or is 6 or greater.

• P(Ac)=1-P(A)

• =1-0.301=0.699

• C) The events A and B are disjoint, so the probability that a first digit is either 1 or is 6 or greater is, by the addition rule,

• P(A or B)=P(A) + P(B)

• =0.301+0.222=0.523

Caution!!!

• Be careful to apply the addition rule only to disjoint events.

D. Check that the probability of the event C that a first digit is odd is

P(C)=P(1) + P(3) +P(5) +P(7) +P(9)=0.609

E. The probability

P(B or C) = ( P(1) + P(3) + P(5) + P(6) + P(9)

= 0.727

Assigning Probabilities:

Equally Likely Outcomes

Random Digits

Applying Probability Rules• The Sample space for a single digit is

« S={1,2,3,4,5,6,7,8,9}

• If all numbers were equally likely then the probability of each outcome would be 1/9.

• The probability of the event B that a random first digit is 6 or greater is

1st  digitor  1 2 3 4 5 6 7 8 9

Probability 1/9

1/9 1/9 1/9 1/9 1/9 1/9 1/91/9

• P(B) = P(6) + P(7)+ P(8)+P(9) =

• 1/9+1/9+1/9+1/9=4/9=0.444

• Compare this with Benford’s law probability. A crook who fakes data by using random digits will end up with too many first digits 6 or greater and too few 1’s and 2’s.

Independence and the Multiplication Rule

« A= first toss is heads

• B= second toss is heads

• P(A) = 0.5, P(B)=0.5 P(A and B) = (.5) (.5) =0.25

• The events (heads on the first toss) and (heads on the second toss) are independent.

Venn Diagram showing the event {A and B}

Which events are Independent?• Tossing of two coins?

• Independent

• Drawing cards without replacement?

• Dependent

• Taking the ACT or SAT more than once?

• dependent

CAUTION!!• The multiplication rule

applies only to independent events; you cannot use it if events are not independent.

Sudden Infant Death

Multiplication RuleOften when more than one SIDs death occurs in a family, the parents are sometimes accussed. One prosecutor claims that there is only a 1 in 72 million chance that two children in the same family could die of SIDS.

The rate of SIDS deaths in a nonsmoking middle class family is 1/8500 so he concludes the probability of two deaths is

1/8500 X 1/8500 = 1/72,250,000

What is wrong with his conclusions?

AIDS testing:Independence and the complement Rule

• A rapid HIV test has a probability of about 0.004 of producing a false-positive. If a clinic tests 200 people who are free of HIV antibodies, what is the probability that one false positive will occur?

• It is reasonable to assume tests on different people are independent. The probability that the test is positive for a single person is 0.004 , so the probability of a negative result is 1-0.004=0.996 by the complement rule.

The probability of at least one false positive among 200 people tested is therefore…

The probability is greater that ½ that at least 1 of the 200 people will test positive for HIV even though no one has the virus.

Assignments• In Class Assignment

Beginning on page 423 6.37, 6.41, 6.44

• Homework Assignment Beginning on page 432 6.53, 6.56, 6.57, 6.58, 6.62

6.2 Summary• Random phenomena: have outcomes that we cannot

predict that nonetheless have a regular distribution in very many repetitions.

• The probability of an event is the proportion of times the event occurs in many repeated trials of a random phenomenon.

• A probability model for a random phenomenon consists of a sample space S and an assignment of probabilities.

• The Sample Space S is the set of all possible outcomes of the random phenomenon.

• Sets of outcomes are called events. A number P(A) is assigned to an event A as its probability.

• A tree diagram begins with a point and draws a line to each possible outcome. Additional lines are drawn to each new outcome, and so on. The result looks like a tree with branches. An outcome in a sample space is one of the paths through the tree.

• The multiplication principle says if you can do one task N number of ways and a second task M # of ways, then both tasks can be done N X M # of ways.

• Sampling with replacement requires that no objects selected from distinct choices be replaced before the next selection.  Probabilities are the same for each draw.  

•  In Sampling without Replacement, probabilities change for each new selection.

• The Complement Ac of an event A consists of exactly the outcomes that are not in A.

• Events A and B are disjoint (mutually exclusive) if they have no outcomes in common.

• Events A and B are Independent if knowing that one event occurs does not change the probability we would assign to the other event.

• Ø is the empty event.  There are no outcomes in the empty event.

• A  Venn Diagram shows events as disjoint or intersecting regions.    

Any assignment of probability must obey the rules that state the basic properties of probability.

• 0≤ P(A) ≤ 1 for any event A.

• P(S) =1 for the sample space S.

• Addition rule: If events A and B are disjoint, then P(A or B) =P(A U B) = P(A) + P(B).

• Complement Rule: For any event A, P(Ac)=1-P(A)

• Multiplication rule: If events A and B are independent, then P(A and B) = P(A B)

= P(A)P(B)

Assignment• P. 356-359

• 6.35, 6.37,6.41,6.45

6.3 General Probability Rules

General Addition Rules

The addition rule for disjoint events P(A or B or C) = P(A) +P(B) +P(C) where events A, B and C are disjoint.

Uniform distribution

Addition rule for disjoint events• Generate a random # X between 0 and 1. What is the probability that the first digit will be odd? The event that the first digit of X is odd is the union of five disjoint events. These events are

The probability that the first digit of a random number is odd

Is the sum of the probabilities of the 5 disjoint events shown.

General Addition Rule for Unions of Two Events: 

P(A or B) =P(A) +P(B) –P(A and B)

Equivalently,

P(AŨB)=P(A) +P(B)­P(A∩B)

Example 6.23• Probability that Deborah gets promoted

P(D)=.7

• Probability that Matt gets promoted

• P(M)=0.5

• Probability that both get promoted =0.3

• P(at least one is promoted)=0.7+0.5-0.3=0.9

• Probability that neither is promoted = 0.1

Joint Event• The simultaneous ocurance of two events, such as (A)

Deborah is promoted and( B )Matt is Promoted.

• The probability of a joint event P(A and B), is called a joint probability

Table for Ex 6.23

Deborah Promoted 0.3 0.7

Not Promoted

Total 0.5

Matthew

Promoted Not PromotedTotal

Completed TableMatthew

 promoted Not promoted Total

Deborah promoted 0.3 0.4 0.7

Not promoted 0.2 0.1 0.3Total 0.5 0.5 1

Assignment• P. 364-365

• 6.47,6.50,6.53

Ex 6.24

Amarillo Slim Wants an Ace• What is the probability that the card dealt to Slim is an Ace?

• P(Ace)=4/52=1/13

• If Slim already has an Ace then his probability of being dealt another Ace becomes

• 3/48 = 1/16

Conditional Probability• The new notation : P(A|B) is Conditional

Probability.

• It gives the probability of one event (A)under the condition that we know another event(B).

Example 6.25

Marital Status of WomenTable 6.1 Shows the Age and Marital of Women (thousands of Women)Table 6.1

Age of Women

18­29 30­64 65+ Total

Married 7,842 43,808 8,270 59,920

Never Married 13,930 7,184 751 21,865

Widowed 36 2,523 8,385 10,944

Divorced

Total

704

22,512

9,174

62,689

1,263

18,699

11,141

103,870

Probability that a Woman is Married

Lets Define two events….

A= The woman chosen is young(18-29)

B= the woman chosen is marriedP(A)=22,512/103,870

0.217

P(A&B)=7842/103,870=

0.075

P(B|A)=7842/22,512=

0.348

P(A and B)=P(A) X P(B|A)The probability that a woman is both young and married is the product of the probabilities that she is young and that she is married given that she is young.

P(A and B)= P(A) X P(B|A)

=22,512/103,870 X 7842/22,512

=7842/103,870

Example 6.27

Slim Wants Diamonds• Slim wants to draw 2 diamonds in a row.

• Between his hand and the upturned cards on the table, there are 11 cards.

• Of these Four are Diamonds.

• The full deck contains 13 diamonds out of 52 cards.

• Therefore, 9 of the 41 remaining cards are Diamonds.

P(second card Diamond|1st Card Diamond) = 8/40

P(both Cards are Diamonds)=

9/41 X8/40

=0.044

P(1st Card is diamond)=9/41

Example 6.28

Finding Conditional Probabilities• What is the conditional probability that a woman is a widow, given that she is at least 65 years old?

• P(at least 65) = 18,669/103,870 =0.180

• P(widowed and at least 65)=8385/103,870=0.081

• The conditional probability is therefore

• P( widowed|at least 65) = .081/0.180 =0.45

Extended Multiplication Rules

Probability of the intersection of Events A, B, and C

P(A and B and C)=

P(A)P(B|A)P(C|A and B)

Example 6.29

The Future of High School Athletes• A= {competes in College}

• B= {competes professionally}

• C= {Pro career longer than 3 years}

P(A)=.05

P(B|A)=0.017

P(C|A and B)= 0.4

P(A and B and C) = P(A)P(B|A)P(C|A and B)

=(.05)x(0.017)x(0.40)

=0.00034

Tree Diagram Revisited

The Probability of reaching the end of any complete branch is the product of the probabilities written on the branches followed.

Conclusions about High School AthletesThere are Two Routes to Professional play

1.Play in college and then go professional

P(B and A)=P(A)P(B|A)=

(0.05)( 0.017) =0.00085

2.Go Straight to Professional from high school

P(B and A^c)=P(A^c)P(B|A^C)

(0.95)(0.0001)=.000095

P(B) =.00085 +.000095=0.000945

About 9 HS athletes in 10,000 will play pro sports

What proportion of professional athletes competed in college?

• P(A|B) =P(A and B)/P(B)

• 0.00085/0.000945=

• 0.8995

• Almost 90 %

Ex 6.30 Online Chat Rooms• 47% of 18-29 yr olds chat online

• 21% of 30-49 yr olds chat online

• 7% of those over 50 chat in online chat rooms

• To determine the % of all internet users that chat online, we need the age breakdown of users.

• 29% of adult internet users are 18-29 (A1),

• 47% are 30-49(A2)

• 24% are over 50 (A3)

What is the probability that a randomly chosen internet user participates in chat rooms?

Bayes’s Rule• If A and B are any events whose probabilities are

not 0 or 1

• P(A|B)=____P(B|A)P(A)__________

> P(B|A)P(A)+P(B|A^c)P(A^c)

Decision Analysis• When making a decision in the presence of uncertainty,

one seeks to make the probability of a favorable outcome as large as possible.

Transplant or Dialysis

The Decision

• The Decision is easy since she has a chance of .857 with the transplant versus only 0.52 without.

Assignment• P. 379-381

• 6.74, 6.76, 6.77

• P. 377-378

• 6.62, 6.63, 6.64

6.3 Summary• The complement Ac of an event A contains all outcomes

that are not in A.

• The union A U B of events A and B contain all outcomes in A, in B or in both A and B.

• The intersection A B contains all outcomes that are in both A and B, but not outomes in A alone or B alone.

The essential general rules for probability are

Legitimate Values: 0≤ P(A) ≤ 1 for any event A.

Total Probability 1: P(S) =1for the sample space S.

General Addition rule:

P(A or B) =P(A U B) = P(A) + P(B) – P(A B)

Complement Rule: For any event A, P(Ac)=1-P(A)

Multiplication rule: P(A B) = P(A)P(B|A)

The conditional probability P(B|A) of an event B, given an event A, is defined by

when P(A)>0. In practice, conditional probabilities are most often found from directly available information.

If A and B are disjoint (mutually exclusive), 

then P (A    B) =0 . 

 The general addition rule for unions then becomes the special addition rule,

 P (A U B)=P(A)+P(B)

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

What you should have Learned• Simulations

• Recognize that many random phenomena can be investigated by means of a carefully designed simulation.

• Use the following steps to construct and run a simulation

• State the problem or describe the random phenomenon

• State the assumption

• Assign digits to represent outcomes

• Simulate many repetitions

• Calculate relative frequencies and state your conclusions

Use a random number table or calculator to conduct simulations

B. Probability Rules• Describe the sample space of a random phenomenon.

For a finite number of outcomes, use the multiplication principle to determine the number of outcomes, and use counting techniques, Venn diagrams and tree diagrams to determine simple probabilities.

For the continuous case, use geometric area to find probabilities (area under simple density curves) of events (intervals on the horizontal axis).

2. Know the probability rules and be able to apply them to determine probabilities of defined events. In particular, determine if a given assignment of probabilities is valid.

3. Determine if two events are disjoint, comlementary, or independent. Find unions and intersections of two or more events.

4. Use Venn diagrams to picture relationships among several events.

5. Use the general addition rule to find probabilities that involve intersecting events.

6. Understand the idea of independence. Judge when it is reasonable to assume independence as part of a probability model

7. Use the multiplication rule for independent events to find the probability that all of several independent events occur.

8. Use the multiplication rule for independent events in combination with other probability rules to find the probability of complex events.

9. Understand the idea of conditional probability.  Find conditional probabilities for  individuals chosen at random from a table of counts of possible outcomes.

10. Use the generral multiplication rule to find the joint probability P(A    B) from P(A) and the conditional probability P(B|A).

11. Construct tree diagrams to organize the use of the multiplication and addition rules to solve problems with several stages.

Assignment

• P. 381

6.70,

p. 385

6.82, 6.83, 6.84, 6.85,