Ch5Probability
Probabilityis a measure of the likelihood of a random phenomenon or chance behavior. Probability describes the long-term proportion with which a certain outcome will occur in situations with short-term uncertainty.
If we flip a coin 100 times and compute the proportion of heads observed after each toss of the coin, what will the proportion approach?
The Law of Large Numbers
As the number of repetitions of a probability experiment increases, the proportion with which a certain outcome is observed gets closer to the probability of the outcome.
Ch 5.1 Probability Rule
Objective A : Sample Spaces and Events
Experiment – any activity that leads to well-defined results calledoutcomes.
Outcome – the result of a single trial of a probability experiment.
Sample space, – the set of all possible outcomes of a probability experiment.
Event, – a subset of the sample space
Simple event, – an event with one outcome is called a simple event.
Compound event – consists of two or more outcomes.
Example 1: A die is tossed one time.
(a) List the elements of the sample space S.
S = {1, 2, 3, 4, 5, 6}
(b) List the elements of this event consisting of a number that is greater than 4.
E = {5, 6}
Example 2: A coin is tossed twice. List the elements of the sample space S, andlist the elements of the event consisting of at least one head.
S = {HH, TT, HT, TH}
e = {HT, TH, HH}
Objective B: Requirements for Probabilities
1. Each probability must lie on between 0 and 1.
2. The sum of the probabilities for all simple events in S equals 1.
If an event is impossible, the probability of the event is 0.
If an event is a certainty, the probability of the event is 1.
An unusual event is an event that has a low probability of occurring. Typically,an event with a
probability less than 0.05 is considered as unusual.
Probabilities should be expressed as reduced fractions or rounded to three decimal places.
Example 1: A probability experiment is conducted. Which of these can beconsidered a probability of an outcome?
(a) 2/5 yes (b) -0.28 can’t be negative (c) 1.09can’t be greater than 1
Example 2: Why is the following not a probability model?
Color / ProbabilityRed / 0.28
Green / 0.56
Yellow / 0.37
No because the sum is 1.21 which is greater than 1
Example 3: Given:
and
Find:
Objective C: Calculating Probabilities
Example 1: The age distribution of employees for this college is shown below:
Age / Number of EmployeesUnder 20 / 25
20 – 29 / 48
30 – 39 / 32
40 – 49 / 15
50 and over / 10
If an employee is selected at random, find the probability that he or she isin the following
age groups
(a) Between 30 and 39 years of age(use 3 decimal places)
P(between 30 and 39) = 32/130 = 0.246
(b) Under 20 or over 49 years of age
P(under 20 or over 49) = (25 + 10)/130 = 35/130 = 0.269
Example 2: Let the sample space be . Suppose the outcomes are equally likely.
(a) Compute the probability of the event .
P(F) = 2/10 = 0.2
(b) Compute the probability of the event = "an odd number."
P(E)= 5/10 = 0.5
Example 3: Two dice are tossed. Find the probability that the sum of two dice is greater than 8?
Sample space: (1,1)(2,1)(3,1)(4,1)(5,1)(6,1)
(1,2)(2,2)(3,2)(4,2)(5,2)(6,2)
(1,3)(2,3)(3,3)(4,3)(5,3)(6,3)
(1,4)(2,4)(3,4)(4,4)(5,4)(6,4)
(1,5)(2,5)(3,5)(4,5)(5,5)(6,5)
(1,6)(2,6)(3,6)(4,6),(5,6)(6,6)
P(sum greater than 8) = 10/36 = 0.278
Example 3: If one card is drawn from a deck, find the probability of getting
(a) a club 13/52 = ¼ = 0.25
(b) a4 and a club1/52 = 0.019
(c) a 4 or a club 16/52 = 4/13 = 0.308
Example 4: Three equally qualified runners, Mark, Bill, and Alan, run a 100-meter sprint, and the
order of finish is recorded.
(a) Give a sample space S.
S = {MBA, MAB, BMA, BAM, AMB, ABM}
(b) What is the probability that Mark will finish last?2/6 = 1/3 = 0.333
Ch5.2 The Addition Rules and Complements
Objective A: Addition Rule for Disjoint (Mutually Exclusive) Events
Event A and B are disjoint (mutually exclusive) if they have no outcomes incommon.
Addition Rule for Disjoint Events
If and are disjoint events, then .
Example 1: A standard deck of cards contains 52 cards. One card is randomlyselected
from the deck. Compute the probability of randomly selecting a two or a three
from a deck of cards.
P(2 or 3) =
Objective B: General Addition Rule
The General Addition Rule
For any two eventsand, .
Example 1: A standard deck of cards contains 52 cards. One card is randomly selected from
the deck. Compute the probability of randomly selecting a two or club from a deck
of cards.
P(2 or club) = 16/52 = 4/13
4 two’s and 13 clubs = 4+13 = 17 subtract 1 card that was counted twice (the 2 of clubs) = 16
Objective C: Complement Rule
Complement Rule
If represents any event and represents the complement of , then.
Example 1: The chance of raining tomorrow is 70%. What is the probabilitythat it will not rain tomorrow?
P(no rain) = 1 – P(rain) = 1 – 0.7 = 0.3 or 30%
Example 2: In a large department store, there are 2 managers, 4 departmentheads, 16 clerks,
and 4 stock persons. If a person is selected at random,
(a)find the probability that the person is a clerk or amanager =
(b) find the probability that the person is not a clerk = 1 – P(clerk) =
Or
Example 3: A probability experiment is conducted in which the sample spaceof the experiment
is .
Let event , event , and event
(a) List the outcome in . Are mutually exclusive?
E and F = {5, 6, 7} not mutually exclusive since there were elements in common
mutually exclusive: no elements in common
(b) Are mutually exclusive? Explain.
F and G = { } yes, no elements in common
(c) List the outcome in . Find by counting the number of outcomes
in . E or F = {2, 3, 5, 6, 7, 8}
P(E or F) out of the sample S : 6/12 = ½ = 0.5
(d) Determine using the General Addition Rule.
P(E) + P(F) – P(E and F) =
(e) List the outcomes in . Find by counting the number of outcomes in .
E={2, 3, 5, 6, 7} so therefore
(f) Determine using the Complement Rule.
Objective D: Contingency Table(for categorical (qualitative) data)
A contingency table relates two categories of data. It is also called a two-way table which consists of
a row variable and a column variable. Each boxinsidethe table is called a cell.
Example 1: In a certain geographic region, newspapers are classified as beingpublished daily morning,
daily evening, and weekly. Some havea comics section and other do not.
The distribution is shown here.
Have comics section / Morning / Evening / Weekly / totalYes / 2 / 3 / 1 / 6
No / 3 / 4 / 2 / 9
Totals / 5 / 7 / 3 / 15 (grand total)
If a newspaper is selected at random, find these probabilities.
(a) The newspaper is a weekly publication.P(weekly) = 3/15 = 0.2
(b) The newspaper is a daily morning publication or has comics.P(morning or comics) = 9/15
(c) P(morning and comics) P(morning and comics) = 2/15
Ch5.3 Independence and the Multiplication Rule
Objective A: Independent Events
Two events are independent if the occurrence of event does not affect theprobability of event .
Two events are dependent if the occurrence of event affects the probabilityof event .
Example 1: Determine whether the events and are independent or dependent. Justify
your answer.
(a) : The battery in your cell phone is dead.
: The battery in your calculator is dead.
independent
(b) : You are late to class.
: You wake up late on a school day.
dependent
Objective B: Multiplication Rule for Independent Events
If and are independent events, then
Example 1: a) A couple wants to have two children. Find the probability that the couple will have two girls.
Sample space: GG BGGBBB
P(G and G) = ¼
Or P(G and G) =
b) A couple wants to have three children. Find the probability that the couple will have three girls.
Sample space: GGG GGBGBGGBB
BGGBGBBBGBBB
P(G and G and G) = 1/8
Or P(G and G and G) =
Example 2: If 36% of college students are underweight, find the probability that if three college
students are selected at random, all will beunderweight.
P(1st and 2nd and 3rd ) = 0.36(0.36)(0.36) = 0.047
Example 3: If 25% of U.S. federal prison inmates are not U.S. citizens,find theprobability that
two randomly selected federal prisoninmates will be U.S. citizens.
P(1st and 2nd) = 0.75(0.75) = 0.563
Objective C: At-Least Probabilities
Probabilities involving the phrase “at least” typically use the Complement Rule. The phrase at
least means “greater than or equal to.” For example, a person must be at least 17 years old to
see an R-rated movie. Means 17 or older.
Example 1: If you make random guesses for two multiple-choice test questions
(each with five possible answers), what is the probability of gettingat least one correct?
Say: 1. A B C D E
2. A B C D E
Method I:
P(at least one correct): It is the probability of getting one correct or two correct = P(one C) or P(two C)=
[ P(1st C and 2nd W) or (1st W and 2nd C)] or [P(1st C and 2nd C)]=
[] + =
Method II:
Can use complement. At least one correct means the complement is none correct
Sample space: CC CWWCWW
P(none correct) = P(1st W and 2nd W) = = 16/25
P(at least one correct) 1 – 16/25 = 9/25
Example 2: According to the Department of Health and Human Services, 30% of 18- to 25-year-olds have
some form of mental illness.
(a) What is the probability two randomly selected 18- to 25-year-olds have some form of
mental illness?
P(1st and 2nd ) = 0.30(0.30) = 0.09
(b) What is the probability six randomly selected 18- to 25-year-olds have some form of
mental illness?
P(1st and 2nd and 3rd and 4th and 5th and 6th) = = 0.07%
(c) What is the probability at least one of six randomly selected 18- to 25-year-olds has
some form of mental illness?
P(at least one) = 1 – P(none) = 1 - = 1 - .117649 = 0.882351 ≈ 0.882
(d) Would it be unusual that among four randomly selected 18- to 25-year-olds, none has
some form of mental illness?
P(1st not and 2nd not and 3rd not and 4th not) = = 24.01%
Since it is more than 5% then it would not be unusual. (It is unusual when the probability of occurring by chance is less than 5%).
Ch 5.4 Conditional Probability and the General Multiplication Rule
Objective A: Conditional Probability and the General Multiplication Rule
A1. Multiplication Rule for Dependent Events
If and are dependent events, then.
The probability of and is the probability of event occurring times the probability of event
occurring, given thatthe eventhas occured.
Example 1: A box has 5 red balls and 2 white balls. If two balls are randomly selected (one after the
other), what is the probability that they both are red?
(a) With replacement(putting first ball back for second selection)
P(1st R and 2nd R) =
(b) Without replacement(not putting the 1st ball back)
P(1st R and 2nd R) =
Example 2: Three cards are drawn from a deck without replacement. Find the probability that all
are jacks.
P(J and J and J) =
A2. Conditional Probability
If and are any two events, then .
The probability of event occurring, given the occurrence of event , isfound by dividing the
probability of and by the probability of .
Example 1: At a local Country Club, 65% of the members play bridge and swim, and 72% play bridge.
If a member is selected at random, find the probabilitythat themember swims, given that
the memberplays bridge.
P(swims l bridge) = = = 0.903
Objective B: Application
Example 1: Eighty students in a school cafeteria were asked if they favored aban on smoking in the
cafeteria. The results of the survey areshown in the table.
Freshman / 15 / 27 / 8 / 50
Sophomore / 23 / 5 / 2 / 30
Totals / 38 / 32 / 10 / 80
If a student is selected at random, find these probabilities.
(a) The student is a freshman or favors the ban.
P(Freshman or favors) = = 73/80 = 0.913
(b) Given that the student favors the ban, the student is asophomore.
P(soph l favors) =
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