1 CHAPTER 7 PROBABILITY, PROBABILITY RULES, AND CONDITIONAL PROBABILITY.

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Transcript of 1 CHAPTER 7 PROBABILITY, PROBABILITY RULES, AND CONDITIONAL PROBABILITY.
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CHAPTER 7
PROBABILITY, PROBABILITY RULES, AND
CONDITIONAL PROBABILITY
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PROBABILITY MODELS: FINITELY MANY OUTCOMES
DEFINITION:
PROBABILITY IS THE STUDY OF RANDOM OR NONDETERMINISTIC EXPERIMENTS. IT MEASURES THE NATURE OF UNCERTAINTY.
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PROBABILISTIC TERMINOLOGIES
• RANDOM EXPERIMENTAN EXPERIMENT IN WHICH ALL OUTCOMES (RESULTS) ARE KNOWN BUT SPECIFIC OBSERVATIONS CANNOT BE KNOWN IN ADVANCE.
EXAMPLE 1:• TOSS A COIN
• ROLL A DIE
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SAMPLE SPACE • THE SET OF ALL POSSIBLE OUTCOMES OF
A RANDOM EXPERIMENT IS CALLED THE SAMPLE SPACE.
• NOTATION: S • EXAMPLE 2
1. FLIP A COIN THREE TIMES
S =
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EXAMPLE 3• AN EXPERIMENT CONSISTS OF FLIPPING A
COIN AND THEN FLIPPING IT A SECOND TIME IF A HEAD OCCURS. OTHERWISE, ROLL A DIE.
• RANDOM VARIABLE
THE OUTCOME OF AN EXPERIMENT IS CALLED A RANDOM VARIABLE. IT CAN ALSO BE DEFINED AS A QUANTITY THAT CAN TAKE ON DIFFERENT VALUES.
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EXAMPLE 4• FLIP A COIN THREE TIMES. IF X
DENOTES THE OUTCOMES OF THE THREE FLIPS, THEN X IS A RANDOM VARIABLE AND THE SAMPLE SPACE IS
• S = {HHH,HHT,HTH,THH,HTT,THT,TTH,TTT}
• IF Y DENOTES THE NUMBER OF HEADS IN THREE FLIPS, THEN Y IS A RANDOM VARIABLE. Y = {0, 1, 2, 3}
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EVENTS
• AN EVENT IS A SUBSET OF A SAMPLE SPACE, THAT IS, A COLLECTION OF OUTCOMES FROM THE SAMPLE SPACE.
• EVENTS ARE DENOTED BY UPPER CASE LETTERS, FOR EXAMPLE, A, B, C, D.
• LET E BE AN EVENT. THEN THE PROBABILITY OF E, DENOTED P(E), IS GIVEN BY
n
iiwPEP
1
)()(
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Probability of an Event, E
•LET E BE ANY EVENT AND S THE SAMPLE SPACE. THE PROBABILITY OF E, DENOTED P(E) IS COMPUTED AS
)(
)()(
Sn
EnEP
Probability of an Event
• The probability of an event A, denoted by P(A), is obtained by adding the probabilities of the individual outcomes in the event.
• When all the possible outcomes are equally likely,
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space sample in the outcomes ofnumber
Aevent in outcomes ofnumber )( AP
Probabilistic Terminologies
• The sample space of a random phenomenon is the set of all possible outcomes. It is usually denoted by S.
• An event is an outcome or a set of outcomes (subset of the sample space).
• A probability model is a mathematical description of longrun regularity consisting of a sample space S and a way of assigning probabilities to events.
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Probability Rules
• Rule 1:– A probability is a number between 0 and 1.
For any event A, 0 ≤ P(A) ≤ 1.– A probability can be interpreted as the
proportion of times that a certain event can be expected to occur.
– If the probability of an event is more than 1, then it will occur more than 100% of the time (Impossible!).
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Probability Rules
• Rule 2:– All possible outcomes together must have
probability 1. P(S) = 1.– Because some outcome must occur on every
trial, the sum of the probabilities for all possible outcomes must be exactly one.
– If the sum of all of the probabilities is less than one or greater than one, then the resulting probability model will be incoherent.
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Probability Rules
• We can use graphs to represent probabilities and probability rules.
• Venn diagram:
Area = ProbabilityP(S) = 1; 0 ≤ P(A) ≤ 1
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S
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EXAMPLE 51. A COIN IS WEIGHTED SO THAT HEADS IS
TWICE AS LIKELY TO APPEAR AS TAILS. FIND P(T) AND P(H).
2. THREE STUDENTS A, B AND C ARE IN A SWIMMING RACE. A AND B HAVE THE SAME PROBABILITY OF WINNING AND EACH IS TWICE AS LIKELY TO WIN AS C. FIND THE PROBABILITY THAT B OR C WINS.
Probability Rules
• Rule 3 (Complement Rule):– The probability that an event does not occur is 1
minus the probability that the event does occur.– The set of outcomes that are not in the event A is
called the complement of A, and is denoted by AC.– P(AC) = 1 – P(A).– Example 6: what is the probability of getting at
least 1 head in the experiment of tossing a fair coin 3 times?
HHH HHT HTH HTTTHH THT TTH TTT
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COMPLEMENT OF AN EVENT
• THE COMPLEMENT OF AN EVENT A WITH RESPECT TO S IS THE SUBSET OF ALL ELEMENTS(OUTCOMES) THAT ARE NOT IN A.
• NOTATION: 'AORAC
)(1)( APAP C
Probability Rules
• Complement rule: P(AC) = 1 – P(A).
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AC
Complement of an Event, A
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Probability Rules
• Rule 4 (Multiplication Rule):– Two events are said to be independent if
the occurrence of one event does not influence the occurrence of the other.
– For two independent events A and B, the probability that both A and B occur is the product of the probabilities of the two events.
– P(A and B) = P(A) X P(B).
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Probability Rules
• A and B in general
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A and B
Probability Rules
• Rule 5 (Addition Rule):– If two events have no outcomes in
common, they are said to be disjoint (or mutually exclusive).
– The probability that one or the other of two disjoint events occurs is the sum of their individual probabilities.
P(A or B) = P(A) + P(B), provided that A and B are disjoint.
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MUTUALLY EXCLUSIVE(DISJOINT) EVENTS
• TWO EVENTS A AND B ARE MUTUALLY EXCLUSIVE(DISJOINT) IF
THAT IS, A AND B HAVE NO OUTCOMES IN COMMON.
IF A AND B ARE DISJOINT(MUTUALLY EXCLUSIVE),
BA
0)()( PBAP
Addition Rule
• Addition rule: P(A or B) = P(A) + P(B), A and B are disjoint.
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A
S
Example 7
• Consider tossing a fair die.– What is the sample space?– Let A be the event that the outcome is odd
and let B be the event that the outcome is a multiple of 3.
• What is (A or B)?• What is (A and B)?• What is the complement of A?• Are A and B disjoint?
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Example 8
• Consider selecting a card from an ordinary deck of 52 cards. Assume the cards are wellshuffled. Let
• A = {the selected is black}• B = {the selected is a picture card}• C = {the selected is an ace}
– What is A and B?– What is B or C?– What is A complement?– Are B and C disjoint?
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Example 9
• The American Red Cross says that about 45% of the U.S. population has Type O blood, 40% Type A, 11% Type B, and the rest Type AB. Someone volunteers to give blood. What is the probability that this donor– has Type AB blood?– has Type A or Type AB?– is not Type O?
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Example 10
• The American Red Cross says that about 45% of the U.S. population has Type O blood, 40% Type A, 11% Type B, and the rest Type AB. Among four potential donors, what is the probability that– all are Type O?– no one is Type AB?– they are not all Type A?– at least one person is Type B?
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Intersection and Union of Events
• INTERSECTION OF EVENTS
THE INTERSECTION OF TWO EVENTS A AND B, DENOTED
IS THE EVENT CONTAINING ALL ELEMENTS(OUTCOMES) THAT ARE COMMON TO A AND B.
BA
Intersection and Union of Events
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UNION OF EVENTS
• THE UNION OF TWO EVENTS A AND B, DENOTED,
IS THE EVENT CONTAINING ALL THE ELEMENTS THAT BELONG TO A OR B OR BOTH.
BA
Probability of the Union of Events
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Probability Rule 6 – The General Addition Rule
• For two events A and B, the probability that A or B occurs equals the probability that A occurs plus the probability that B occurs minus the probability that A and B both occur.
P(A or B) = P(A) + P(B) – P(A and B).
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The General Addition Rule
• A ={students in Spanish class} and B = {students in French class}
• “A or B = • A + B – (A and B)”
–A or B in terms of color: Blue, Red, Yellow–A in terms of color: Red, Yellow–B in terms of color: Blue, Red–A and B in terms of color: Red
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Example 11
• An elementary school is offering 2 language classes: one in Spanish, and one in French. There are 60% students in the Spanish class, 56% in the French class, and 30% students in both language classes.– What percent of students are in either Spanish
class or French class or both?– If one student is randomly selected from this
school, what is the probability that she/he is in either Spanish class or French class but not both?
– For a randomly selected student, what is the probability that she/he is only in Spanish class?
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Example 12
• Suppose the probability that a U.S. resident has traveled to Canada is 0.18, to Mexico is 0.09, and to both countries is 0.04. What is the probability that a U.S. resident chosen at random has– a) traveled to either Canada or Mexico or
both?– b) traveled to Canada but not Mexico?– c) not traveled to either country?
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Example 12 Cont’d• C ={traveled to Canada} and M = {traveled to Mexico}
– Blue: to Mexico but not Canada– Red: to both Canada and Mexico0.04– Yellow: to Canada but not Mexico– Blue and Red: to Mexico0.09– Yellow and Red: to Canada0.18
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NULL EVENT: AN EVENT THAT HAS NO CHANCE OF OCCURING. THE PROBABILITY OF A NULL EVENT IS ZERO. P( NULL EVENT ) = 0
• CERTAIN OR SURE EVENT: AN EVENT THAT IS SURE TO OCCUR. THE PROBABILITY OF A SURE OR CERTAIN EVENT IS ONE.
P(S) = 1
Conditional Probability
• The probability that the event B occurs given that the event A occurs is called the conditional probability of the event B given the event A, and is denoted by P(BA).
• Moreover,
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P(A)
B) andP(A A)P(B
Conditional Probability
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Conditional Probability
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A and B
Example 13
• If a young girl is randomly selected from all girls surveyed during 8788, 8991, and 9496, what is the probability that– she reports drinking milk? – she reports drinking milk and she is from the 8991 survey ?– she reports drinking milk if we know she is from the 8991
survey?41
Nationwide Survey Years
8788 8991 9496 Total
Drink Milk
Yes 354 502 366 1222
No 226 335 336 927
Total 580 837 732 2149
Example 13 Cont’d
• For a randomly selected girl, the probability that – she reports drinking milk and she is
from the 8991 survey is 502/2149;– she is from the 8991 survey is
837/2149;– she reports drinking milk given that
she is from the 8991 survey is (502/2149)/(837/2149) = 502/837.
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Example 14
• When choosing two cards from a deck of wellshuffled cards, what is the probability that the second card is black?
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GENERAL MULTIPLICATION RULE• THE FORMULA FOR CONDITIONAL PROBABILITY
CAN BE MANIPULATED ALGEBRAICALLY SO THAT THE JOINT PROBABILITY P(A and B) CAN BE DETERMINED FROM THE CONDITIONAL PROBABILITY OF AN EVENT. USING
AND SOLVING FOR P(A and B), WE OBTAIN THE GENERAL MULTIPLICATION RULE
)(
)()(
BP
BandAPBAP
)().()( BPBAPBandAP
CONDITIONAL PROBABILITY
BAYE’S FORMULA THROUGH
TREE DIAGRAMS
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BAYES’ FORMULA FOR TWO EVENTS A AND B
• BY THE DEFINITION OF CONDITIONAL PROBABILITY,
)(
)().(
)(
)()(
AP
BPBAP
AP
BandAPABP
)().()().(
)().(CC BPBAPBPBAP
BPBAP
Examples From Midterm II Practice Sheet
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Summary
• Law of large numbers
• Probability: Sample space / Events
• Rules for probability model:
1. for any event A, 0 ≤ P(A) ≤ 1.
2. for sample space S, P(S) = 1.
3. if two events A and B are disjoint, then P(A or B) = P(A) + P(B).
4. for any event A, P(A does not occur) = 1  P(A).
5. For two independent events A and B, P(A and B) = P(A) X P(B).
• Venn diagram48
Summary of Chapter
• General Addition Rule:– For two events A and B,
P(A or B) = P(A) + P(B) – P(A and B).• General Multiplication Rule
– For two events A and B, P(A and B) = P(BA) X P(A).
• Conditional probability
• Independence: P(BA) = P(B).
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P(A)
B) andP(A A)P(B
Example 15 – Self Read
• 56% of all American workers have a workplace retirement plan, 66% have health insurance, and 73% have at least one of the benefits. We select a worker at random.– What is the probability that he has both health
insurance and a retirement plan?– What is the probability that he has neither health
insurance nor a retirement plan?– What is the probability that he only has a
retirement plan?– Knowing that he has a retirement plan, what is the
probability that he has health insurance?
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Example 15 – Self Read
• Solution:– Let A be the event that he has a retirement plan.– Let B be the event that he has health insurance.– Then P(A)=0.56, P(B)=0.66, and P(A or B)=0.73.
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Example 15 – Self Read
• Solution:– What is the probability that he has both health insurance
and a retirement plan?• P(A and B)=?• General addition rule: P(A
or B) = P(A) + P(B)  P(A and B)• Therefore, P(A and B) = P(A) + P(B)  P(A or B) =
0.56+0.660.73 = 0.49
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Example 15 – Self Read
• Solution:– What is the probability that he has neither health insurance
nor a retirement plan?• The probability that he has at least one benefit is 0.73.• Therefore, the probability that he has neither health
insurance nor a retirement plan is 10.73=0.27.
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Example 15 – Self Read
• Solution:– What is the probability that he only has a retirement plan?
• “Only has a retirement plan” means has a retirement plan but no health insurance (not both).
• Therefore, • P(he only has a retirement plan) = P(A) – P(A and B) =
0.560.49 = 0.07
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Example 15 – Self Read
• Solution:– Knowing that he has a retirement plan,
what is the probability that he has health insurance?
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.875.056.0
49.0
P(A)
A) and P(BA)P(B
ExtraCredit Exercise
• In a large Statistics lecture, the professor reports that 52% of the students enrolled have never taken a Calculus course, 34% have taken only one semester of Calculus, and the rest have taken two or more semesters of Calculus. The professor randomly assigns students to groups of three to work on a project for the course.– What is the probability that the first group member you
meet has studied some Calculus?– What is the probability that the first group member you
meet has studied no more than one semester of Calculus?– What is the probability that both of your two group
members have studied exactly one semester of Calculus?– What is the probability that at least one of your group
members has had more than one semester of Calculus?
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ExtraCredit Exercise
• A North American roulette wheel has 38 slots, of which 18 are red, 18 are black, and 2 are green. If you bet on red, the probability of winning is 18/38 = .4737. The probability .4737 represents
(A) nothing important, since every spin of the wheel results in one of three outcomes (red, black, or green).(B) the proportion of times this event will occur in a very long series of individual bets on red.(C) the fact that you're more likely to win betting on red than you are to lose.(D) the fact that if you make 100 wagers on red, you'll have 47 or 48 wins.
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Extra Credit Problems
1. A PAIR OF FAIR DICE IS TOSSED. FIND THE PROBABILITY THAT THE MAXIMUM OF THE TWO NUMBERS IS GREATER THAN 4.
2. ONE CARD IS SELECTED AT RANDOM FROM 50 CARDS NUMBERED 1 TO 50. FIND THE PROBABILITY THAT THE NUMBER ON THE CARD IS (I) DIVISIBLE BY 5, (II) PRIME, (III) ENDS IN THE DIGIT 2.