Probability and Sampling
Each day decisions are based
Should you buy an extended warranty
for your new digital camera?
Should you allow 25 min to get school
or is 15 min enough?
If an artificial heart has four key parts,
how likely is each one to fail? How
likely is it that at least one will fail?
Chance experiment – any activity or situation
in which there is uncertainty about which of
two or more possible outcomes will result.
Chance behavior is unpredictable in the short
run but has a regular and predictable pattern
in the long run
Sample space – the collection of all possible
outcomes of a chance experiment.
Sample space of chance
Experiment - whether men and
women have different shopping
preferences when buying a CD
at a music store – classical,
rock, country, and “other”.
Tree diagram of sample space:
Event – any collection of outcomes from
the sample space of a chance
Simple event – an event consisting of
exactly one outcome
Probability of an event E
P(E) = number of outcomes favorable to E
number of outcomes in the sample space
***only when the outcomes of an experiment are
equally likely – fair coins or dice, etc.
Any probability is a number between 0
All possible outcomes together must
have a probability of 1
The probability that an event does not
occur is 1 minus the probability that the
event does occur
A mathematical description of a random
phenomenon consisting of a sample
space and a way of assigning
probabilities to events.
Construct a probability model for a
family with 3 children.
All probabilities must add to one
Pg 226 #4.19, 4.20, 4.21
1. What is the probability of rolling a 3 with one die?
2. What is the probability of picking a Queen in a deck
4/52 = 1/13
3. What is the probability of picking a heart in a deck
4. What is the probability of rolling a sum of 8 with
(2,6)(6,2)(3,5)(5,3)(4,4) = 5/36
5. What is the probability of a family with three children to have all
6. A survey was taken of 10,672 families to determine the number
of televisions owned by each. The following results were
Number of Televisions Owned Frequency
Find the probability of a person having:
a. Two televisions
b. Between one and three televisions, inclusive
7270/10672 = 3635/5336
c. Seven televisions
Ways of finding # of
Permutations – in a line
Arrangement in line with duplicates
Arrangement in circle
Combinations – in a group
If one thing can be done in m ways and if after this is done,
something else con be done in n ways, then both things can be
done in a total of (m)(n) different ways in the stated order.
1. A certain model car comes with one of three possible engine sizes
and with or without an AM/FM radio. Furthermore, it is equipped
with automatic or standard transmission. In how many different
ways can a buyer select a car?
3 · 2 · 2 = 12
2. There are nine approach roads leading to an airport. Because of
heavy traffic, a taxi driver decides to go to the airport by one road
and to leave by another road. In how many different ways can this
9 · 8 = 72
How many different numbers greater than 3000 can be formed from
the digits 2, 3, 5, and 9 if no repetitions are allowed?
3 · 3 · 2 · 1 = 18
Arrangement of distinct objects in a particular order
Equation: nPr or n!
(n r )!
1. In a supermarket there is a long line at the checkout counter. The
manager notices this and decides to open an additional checkout
counter. Seven people rush over to the new checkout counter. In
how many different ways can these seven people line up to be
7! = 5040
2. Susan is an IRS agent. She has made appointments with eight
taxpayers to review their 1040 tax forms on May 2 on a first-come,
first-served basis. However, due to a computer malfunction, she
finds that she has time to meet with only five taxpayers to review
their forms. Assuming order counts, in how many different ways
can this be done?
8P5 or 8 · 7 · 6 · 5 · 4 = 6720
Each year movie-goers in a certain city are asked to rank the 5 best
movies from among a list of 14 movies. In how many different ways
can this be done?
14P5 or 14 · 13 · 12 · 11 · 10 = 240240
The number of different permutations of n
things of which p are alike, q are alike, or r are
How many different ways can we
arrange the letters in the word
Arrangement of distinct
objects in a circle
Equation: (n – 1)!
How many different ways can four
people be arranged in a circle?
3! = 6
A selection from a collection of distinct objects where order is not
Equation: nCr or r!(n r )!
1. Medical researchers are testing a new drug for treating one form of
a neurological disorder. It is decided to select a random group of
18 people and then to select 8 of these people to be given the new
drug. The remaining 10 people will be given a placebo. In how
many different ways can the 8 subjects be selected?
18C8 = 43758
2. There are ten nurses who work on the night shift on the tenth floor
of General Hospital. In an effort to save money, the hospital
administrator decides to fire four nurses. In how many different
ways can the administrator select the four nurses to be fired?
10C4 = 210
The odds in favor of an event occurring are p to q,
where p is the number of favorable outcomes and
q is the number of unfavorable outcomes.
1. Find the odds against rolling a 5 when a single die
is rolled once.
2. A roulette wheel has 38 slots. One slot is 1,
another is 00, and the others are numbered 1
through 36, respectfully. You are placing a bet that
the outcome is an odd number. What are the odds
18 : 38
3. What are the odds against winning?
20 : 38
Probability with addition
To find the probability of one event or the other to
occur, add the probabilities of each one happening
1. Two dice are rolled. What is the probability that the
sum of the dots appearing on both dice together is 9 or
4/36 + 2/36 = 6/36 = 1/6
2. One card is drawn from a deck of cards. What is the
probability of getting a king or a red card?
4/52 + 26/52 – 2/52 = 28/52
If two events are independent events then the
probability of both things happening is P(A) · P(B)
1. Philip, Janet, and Fredric have each applied to different
banks for a home equity loan. The probability that
Philip’s application is approved is .85. The probability
that Janet’s application is approved is .92, and the
probability that Fredric’s application is approved is .79.
Assuming independence, find the probability that all
three applications are approved.
(.85)(.92)(.79) = .61778
2. From the problem above find the probability that none
of the applications are approved.
(.15)(.08)(.21) = .00252