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CHAPTER 12
PROBABILITY
12.1 Basic Concepts
12.2 Events Involving “Not” and “Or”
12.3 Conditional Probability; Events Involving “And”
12.1 Basic Probability Concepts
The study of probability is concerned with such random phenomena. Even though we
cannot be certain whether a given result will occur, we often can obtain a good
measure of its likelihood, or probability. This chapter discusses various ways of finding
and using probabilities.
The Language of probability
Experiment: Any observation, or measurement, of a random phenomenon,
Outcomes: The possible results of the experiment,
Sample space: the set of all possible outcomes,
Event: Any such subset of the sample space.
Let’s Do it!
Find the Sample space of the following experiments:
A) The experiment of rolling a fair 8-sided die.
B) The experiment of drawing a card and noting the suite of the card.
C) Tossing two unbiased coins.
D) Gender of children in a 3-child family.
Finding Probability of an outcome
Example Finding the Probability of Having Daughters
Michelle Brown wants to have exactly two daughters. Assuming that boy and girl
babies are equally likely, find her probability of success if
(a) She has a total of two children,
(b) She has a total of three children.
SOLUTION
a) The sample space S={gg, gb, bg, bb}. Only one outcome is favorable to the event
E={gg} of exactly two daughters.  P(E ) 
n(E ) 1

n(S) 4
b) The sample Space S={ ggg, ggb, gbg, gbb, bgg, bgb, bbg, bbb} and E={ggb, gbg, bgg},
3
8
so P(E)  .
Let’s Do it! Number Sums for Rolling Two Dice
Two 6-sided dice are rolled
a) Find the sample space S.
b) Find the probability of rolling a sum of 2.
c) Find the probability of rolling a sum of 7.
Find the probability of rolling a sum of 11.
Example
1) What is the size of the sample space of a 5-hand poker (i.e. How many possible
hands in a 5-card poker).
The sample space is the set of all possible 5-card hands. That is 52 C 5 = 2,598,960.
2) Find the probability of being dealt each of the following hands in five-card poker.
Use a calculator to obtain answers to eight decimal places.
a) A royal Flush (Ace, king, queen, jack, and 10, all of the same suit)
There are 4 possible royal-flush hands. Therefore,
P(royal flush)= 4 2,598,960
=1.539077169x10-6
=0.00000154
b) Two pairs :(2 cards of one denomination, 2 cards of a second denomination,
and 1 card of a third denomination.)
To find the number hands that are two pairs, we consider this as a multi-task process.
First: there are 13 different denominations and we are interested in picking 2
denominations for the pairs (that is 13 C 2 ), that leaves 11 denominations for the single
card (that is 11 C1 ). . Second: Each denomination has 4 suits to choose for two pairs and
a single card. Therefore,
Number of possible twp pairs is= 13C2 x 4C2 x 4C2 x 11C1 x 4C1= 123,552
P(Two Pairs)= 123,552
2,598,960
=0.04753902
Let’s Do it! Poker (Level of difficulty=difficult)
Find the probability of dealing the following hands in five-card poker. Use a calculator
to obtain answers to eight decimal places.
a) A royal Flush (Ace, king, queen, jack, and 10, all of the same suit)
b) A full house (three of one denomination and 2 of another)
Probability in Genetics
An important early example was the work of the Austrian monk Gregor Mendel, who
used the idea of randomness to help establish the study of genetics. In an effort to
understand the mechanism of character transmittal from one generation to the next in
plants, Mendel counted the number of occurrences of various characteristics.
For example, he found that the flower color in certain pea plants obeyed this scheme:
Pure red crossed with pure white produces red.
Mendel theorized that red is “dominant” (symbolized with the capital letter R), while
white is “recessive” (symbolized with the lowercase letter r).
The pure red parent carried only genes for red (R), and the
pure white parent carried only genes for white (r).
Now each of these second-generation offspring, though
exhibiting the color red, still carries one of each gene.
When two of second generations are crossed, each thirdgeneration offspring will receive one of the gene
combinations shown in Table
Example
Referring to Table above, determine the probability that a third-generation offspring
will exhibit each of the following flower colors. Base the probabilities on the following
sample space of equally likely outcomes:
a) Red
b) White
Solution
a) Since red dominates white, any combination with at least one gene for red (R) will
result in red flowers. Since three of the four possibilities meet this criterion,
P(Re d) 
3
.
4
b) Only the combination rr has no gene for red, so, P(White) 
1
.
4
Let’s Do It!
Mendel also investigated various characteristics besides flower color. For example,
round peas are dominant over recessive wrinkled peas. First, second, and third
generations can again be analyzed using Tables, where R represents round and r
represents wrinkled.
a) Explain why crossing pure round and pure wrinkled first-generation parents will
always produce round peas in the second-generation offspring.
b) When second-generation round pea plants (each of which carries both R and r
genes) are crossed, find the probability that a third-generation offspring will have
(i) Round peas
(ii) Wrinkled peas.
Odds
Odds: Compare the number of favorable outcomes to the number of unfavorable
outcomes.
Odds are commonly quoted, in horse racing, lotteries, and most other gambling
situations. And the odds quoted normally are odds “against” rather than odds “in
favor.”
Example
Jennifer has been promised one of six summer jobs, three of which would be at a
nearby beach resort. If she has equal chances for all six jobs, find the odds that she will
land one at the resort.
Solution
Since three possibilities are favorable and three are not, the odds of working at the
resort are 3 to 3, or 1 to 1. (The common factor of 3 has been divided out.)
Odds of 1 to 1 are often termed “even odds.”
Let’s Do It!
Donn Demaree has purchased six tickets for an office raffle where the winner will
receive an iPod unit. If 51 tickets were sold altogether and each has an equal chance of
winning, what are the odds against Donn’s winning the iPod?
Let’s Do It! Using Spinners to Generate Numbers
Suppose the spinner shown here is spun once, to determine a singledigit number, and we are interested in the event E that the resulting
number is odd. Give each of the following.
(a) The sample space
(b) The number of favorable outcomes
(c) The number of unfavorable outcomes
(e) The probability of an odd number
(f) The odds in favor of an odd number
End of section 12.1. Start your online homework on MyMathLab.
12.2 Events Involving “Not” and “Or”
Properties of Probability
Let’s Do It!
When a single fair die is rolled, find the probability of each event.
(a) The number 2 is rolled
(b) a number other than 2 is rolled
(c) The number 7 is rolled
(d) a number less than 7 is rolled
Events Involving “Not”
Understanding Complements
Example
When rolling a single fair six-sided die once. The sample space is S={1,2,3,4,5,6}. Let event
E=rolling a number less than 5.
P(E)=4/6
The complement of E is denoted by E’ ={ rolling a 5 or more}
P (E’) =1-P(E) =2/6
Let’s Do It!
When a single card is drawn from a standard 52-card deck, what is the probability that it will
not be a king?
Events Involving “Or”
Example
If one number is selected randomly from the set {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} find the
probability that it will be
a) Odd or a multiple of 4 b) Odd or a multiple of 3.
Solution
Sample Space S ={1, 2, 3, 4, 5, 6, 7, 8, 9, 10}
A=set of odd values = {1, 3, 5, 7, 9}
B= set of multiples of 4 = {4, 8}
C=set of multiples of 3 = { 3, 6, 9}
a)
P(odd or multiple 4)=P(Aor B)=7/10
b)
P(odd or multiple of 3)=P(A or C)=6/10.
Let’s Do It!
If a single card is drawn from a standard 52-card deck, what is the probability that it will be a
spade or a red card?
Find the probability that a single card drawn from a standard 52-card deck will be a diamond
or a face card.
Of 20 elective courses to be offered this term, Juanita plans to enroll in one, which she will
choose by throwing a dart at the schedule of courses. If 8 of the courses are recreational,
9 are interesting, and 3 are both recreational and interesting, find the probability that the
course she chooses will have at least one of these two attributes.
End of section 12.2. Start your online homework on MyMathLab.
12.3 Conditional Probability; Events Involving “And”
Conditional Probability Sometimes the probability of an event must be computed using the
knowledge that some other event has happened (or is happening, or will happen—the timing
is not important). This type of probability is called conditional probability.
Example
From the sample space S= {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} a single number is to be selected
randomly. Given the events
A: The selected number is odd, and B: The selected number is a multiple of 3
Find each probability.
a) P (A)
A= {1, 3, 5, 7, 9}  P(A) = 5/10 = ½
b) P (B)
B = {3, 6, 9}  P(B) = 3/10
c) Probability that the number selected is odd and also a multiple of 3.
(A and B) = {3, 9}  P(A and B) = 2/10 = 1/5
d) Probability that a selected number is a multiple of 3 given that it is an odd number.
The given condition, that A occurs, effectively reduces the sample space from S to A, and the
elements of the new sample space A, that are also in B, are the elements of ( A and B) .
Therefore,
2
P(B given A) 
5
Let’s Do it!
Given a family with two children, Define the following events:
A= both children are girls, B=At least one child is a girl, C= the older one is a girl.
Find the probabilities that
(a) Both are girls, given that at least one is a girl.
(b) Both are girls, given that the older child is a girl.
Let’s Do It!
Each year, Diane Carr adds to her book collection a number of new publications that
She believes will be of lasting value and interest. She has categorized each of her twenty
acquisitions for 2007 as hardcover or paperback and as fiction or nonfiction. The numbers of
books in the various categories are shown in the Table
If Diane randomly chooses one of these 20 books, find the probability it will be
(a) Hardcover,
(b) Hardcover fiction
(c) Fiction, given it is hardcover
Independent Events
Example
Table lists the nine generally acknowledged planets of our solar system (prior
to 2006) together with their mean distances from the sun, in millions of
kilometers.(Data is from the Time Almanac 1999.).
Selina must select two planets, one is for an oral report, the other is for a
written report, and they need not be distinct. (The same planet may be
selected for both.) .
Find the probability that, if she selects randomly, the first is closer than Mars
and the second is closer than Saturn.
Solution
We define the events
A: The first is closer than Mars,
and
B: The second is closer than Saturn.
Since the selection is done with replacement. Repetitions are allowed. In this case event B is
independent of event A. So,
P( A and B)  P( A)  P(B) = 1/3. 5/9 = 5/27
Let’s Do It! Comparing Gender and Career Motivation of University Students
One hundred college seniors attending a career fair at a major northeastern university were
categorized according to gender and according to primary career motivation, as summarized
here
If one of these students is to be selected at random, find the probability that the student
selected
will satisfy each condition.
A) Female
B) Motivated primarily by creativity
C) Not motivated primarily by money
D) Male and motivated primarily by money
E) Are gender and motivation independent? Why?
E) Male, given that primary motivation is a sense of giving to society
F) Motivated primarily by money or creativity, given that the student is female
.
End of section 12.3. Start your online homework on MyMathLab.