Transcript College Algebra Fifth Edition James Stewart  Lothar Redlin  Saleem Watson Counting and Probability.

College Algebra
Fifth Edition
James Stewart  Lothar Redlin

Saleem Watson
10
Counting
and Probability
10.3 Probability
Overview
In the preceding chapters, we
modeled real-world situations.
• These were modeled using precise
rules, such equations or functions.
Overview
However, many of our everyday
activities are not governed by
precise rules.
• Rather, they involve
randomness and uncertainty.
Overview
How can we model such situations?
• Also, how can we find reliable
patterns in random events?
• In this section, we will see how
the ideas of probability provide
answers to these questions.
Rolling a Die
Let’s look at a simple example.
• We roll a die, and we’re hoping to get
a “two”.
• Of course, it’s impossible to predict what
number will show up.
Rolling a Die
But, here’s the key idea:
• We roll the die many many times.
• Then, the number two will show up
about one-sixth of the time.
Rolling a Die
This is because each of the six
numbers is equally likely to show
up.
• So, the “two” will show up about
a sixth of the time.
• If you try this experiment, you will
see that it actually works!
Rolling a Die
We say that the probability
(or chance) of getting “two”
is 1/6.
Picking a Card
If we pick a card from a 52-card deck,
what are the chances that it is an ace?
• Again, each card is equally likely to be
picked.
• Since there are four aces, the probability
(or chances) of picking an ace is 4/52.
Probability and Science
Probability plays a key role in many
sciences.
• A remarkable example of the use of
probability is Gregor Mendel’s discovery
of genes.
• He could not see the genes.
• His discovery was due to applying
probabilistic reasoning to the patterns
he saw in inherited traits.
Probability
Today, probability is an indispensable
tool for decision making.
• It is used in business, industry, government,
and scientific research.
• For example, probability is used to
–Determine the effectiveness of new medicine
–Assess fair prices for insurance policies
–Gauge public opinion on a topic
(without interviewing everyone)
Probability
In the remaining sections of this
chapter, we will see how some of
these applications are possible.
What is Probability?
Terminology
To discuss probability, let’s begin by defining
some terms.
• An experiment is a process, such as tossing a
coin or rolling a die.
• The experiment gives definite results called the
outcomes of the experiment.
– For tossing a coin, the possible outcomes are
“heads” and “tails”
– For rolling a die, the outcomes are 1, 2, 3, 4, 5,
and 6.
Terminology
The sample space of an experiment is the
set of all possible outcomes.
• If we let H stand for heads and T for tails,
then the sample space of the coin-tossing
experiment is S = {H, T}.
Sample Space
The table lists some experiments and the
corresponding sample spaces.
Experiments with Equally Likely Outcomes
We will be concerned only with experiments
for which all the outcomes are equally likely.
• We already have an intuitive feeling for what
this means.
• When we toss a perfectly balanced coin,
heads and tails are equally likely outcomes.
• This is in the sense, that if this experiment
is repeated many times, we expect that about
as many heads as tails will show up.
Experiments and Outcomes
In any given experiment, we are often
concerned with a particular set of outcomes.
• We might be interested in a die showing an
even number.
• Or, we might be interested in picking an ace
from a deck of cards.
• Any particular set of outcomes is a subset of
the sample space.
An Event—Definition
This leads to the following definition.
• If S is the sample space of an experiment,
then an event is any subset of the sample
space.
E.g. 1—Events in a Sample Space
An experiment consists of tossing a coin
three times and recording the results in order.
•
The sample space is
S = {HHH, HHT, HTH, THH,
TTH, THT, HTT, TTT}
E.g. 1—Events in a Sample Space
The event E of showing “exactly two heads” is
the subset of S.
•
E consists of all outcomes with two heads.
•
Thus,
E = {HHT, HTH, THH}
E.g. 1—Events in a Sample Space
The event F of showing “at least two heads”
is
F = {HHH, HHT, HTH, THH}
And, the event of showing “no heads” is
G = {TTT}
Intuitive Notion of Probability
We are now ready to define the notion of
probability.
• Intuitively, we know that rolling a die may
result in any of six equally likely outcomes.
• So, the chance of any particular outcome
occurring is 1/6.
Intuitive Notion of Probability
What is the chance of showing an even
number?
• Of the six equally likely outcomes possible,
three are even numbers.
• So it is reasonable to say that the chance of
showing an even number is 3/6 = 1/2.
• This reasoning is the intuitive basis for the
following definition of probability.
Probability—Definition
Let S be the sample space of an experiment
in which all outcomes are equally likely.
• Let E be an event.
• The probability of E, written P(E), is
n(E ) number of elements in E
P (E ) 

n(S ) number of elements in S
Values of a Probability
Notice that 0 ≤ n(E) ≤ n(S).
• So, the probability P(E) of an event is a number
between 0 and 1.
• That is, 0 ≤ P(E) ≤ 1.
• The closer the probability is to 1, the more likely
the event is to happen.
• The close to 0, the less likely.
Values of a Probability
If P(E) = 1, then E is called the certain event.
• If P(E) = 0, then E is called the impossible event.
E.g. 2—Finding the Probability of an Event
A coin is tossed three times, and the results
are recorded.
•
What is the probability of getting exactly
two heads?
•
At least two heads?
•
No heads?
E.g. 2—Finding the Probability of an Event
By the results of Example 1, the sample
space S of this experiment contains eight
outcomes.
•
The event E of getting “exactly two heads”
contains three outcomes.
•
They are {HHT, HTH, THH}.
•
So, by the definition of probability,
n( E ) 3
P (E ) 

n(S ) 8
E.g. 2—Finding the Probability of an Event
Similarly, the event F of getting “at least two
heads” has four outcomes.
•
They are {HHH, HHT, HTH, THH}.
•
So,
n(E ) 4 1
P (E ) 
 
n(S ) 8 2
E.g. 2—Finding the Probability of an Event
The event G of getting “no heads” has one
element, so
n( E ) 1
P (E ) 

n(S ) 8
Calculating Probability
by Counting
Calculating Probability by Counting
To find the probability of an event:
• We do not need to list all the elements in
the sample space and the event.
• What we do need is the number of elements
in these sets.
• The counting techniques that we learned in
the preceding sections will be very useful
here.
E.g. 3—Finding the Probability of an Event
A five-card poker hand is drawn from a
standard 52-card deck.
•
What is the probability that all five cards
are spades?
•
The experiment here consists of choosing
five cards from the deck.
•
The sample space S consists of all possible
five-card hands.
E.g. 3—Finding the Probability of an Event
Thus, the number of elements in the sample
space is
n(S )  C(52,5)
52!

5!(52  5)!
 2,598,960
E.g. 3—Finding the Probability of an Event
The event E that we are interested in consists
of choosing five spades.
•
Since the deck contains only 13 spades,
the number of ways of choosing five spades
is
n(E )  C(13,5)
13!

5!(13  5)!
 1,287
E.g. 3—Finding the Probability of an Event
Thus, the probability of drawing five spades is
n(E )
P (E ) 
N (S )
1,287

2,598,960
 0.0005
Understanding a Probability
What does the answer to Example 3 tell us?
• Since 0.0005 = 1/2000, this means that if you
play poker many, many times, on average
you will be dealt a hand consisting of only
spades about once every 2000 hands.
E.g. 4—Finding the Probability of an Event
A bag contains 20 tennis balls.
•
Four of the balls are defective.
•
If two balls are selected at random from
the bag, what is the probability that both
are defective?
E.g. 4—Finding the Probability of an Event
The experiment consists of choosing two
balls from 20.
•
So, the number of elements in the sample
space S is C(20, 2).
•
Since there are four defective balls,
the number of ways of picking two
defective balls is C(4, 2).
E.g. 4—Finding the Probability of an Event
Thus, the probability of the event E of picking
two defective balls is
n( E )
P (E ) 
n(S )
C (4,2)

C (20,2)
6

20
 0.032
Complement of an Event
The complement of an event E is the set
of outcomes in the sample space that is
not in E.
• We denote the complement of an event E
by E′.
Complement of an Event
We can calculate the probability of E′ using
the definition and the fact that
n(E′) = n(S) – n(E)
• So, we have
n(E ') n(S )  n(E ) n(S ) n(E )
P (E ') 



n(S )
n(S )
n(S ) n(S )
 1  P (E )
Probability of the Complement of an Event
Let S be the sample space of an experiment,
and E and event.
• Then
P(E ')  1  P(E )
Probability of the Complement of an Event
This is an extremely useful result.
• It is often difficult to calculate the probability
of an event E.
• But, it is easy to find the probability of E′.
• From this, P(E) can be calculated immediately
by using this formula.
E.g. 5—The Probability of the Complement of an Event
An urn contains 10 red balls and 15 blue
balls.
•
Six balls are drawn at random from the urn.
•
What is the probability that at least one ball
is red?
E.g. 5—The Probability of the Complement of an Event
Let E be the event that at least one red ball is
drawn.
•
It is tedious to count all the possible ways
in which one or more of the balls drawn
are red.
•
So let’s consider E′, the complement of this
event.
•
E′ is the event that none of the balls drawn
are red.
E.g. 5—The Probability of the Complement of an Event
The number of ways of choosing 6 blue balls
from the 15 balls is C(15, 6).
•
The number of ways of choosing 6 balls
from the 25 ball is C(25, 6).
•
Thus,
n(E ') C(15,6)
5,005
P (E ') 


n(S ) C(25,6) 177,100
13

460
E.g. 5—The Probability of the Complement of an Event
By the formula for the complement of an
event, we have
P (E )  1  P (E ')
13
 1
460
447

460
 0.97
The Union of Events
The Union of Events
If E and F are events, what is the probability
that E or F occurs?
• The word or indicates that we want the probability
of the union of these events.
• That is, E  F .
The Union of Events
So, we need to find the number of elements
in E  F.
• If we simply add the number of element in E
to the number of elements in F, then we would
be counting the elements in the overlap twice.
• Once in E and once in F.
The Union of Events
So to get the correct total, we must subtract
the number of elements in E  F.
• Thus,
n(E  F )  n(E )  n(F )  n(E  F )
The Union of Events
Using the formula for probability,
we get
n( E  F ) n ( E )  n( F )  n( E  F )
P (E  F ) 

n(S )
n(S )
n(E ) n(F ) n(E  F )



n(S ) n(S )
n(S )
 P (E )  P (F )  P (E  F )
• We have just proved the following.
The Probability of The Union of Events
If E and F are events in a sample space S,
then the probability of E or F is
P(E  F )  P(E )  P(F )  P(E  F )
E.g. 6—The Probability of the Union of an Event
What is the probability that a card drawn at
random from a standard 52-card deck is
either a face card or a spade?
•
We let E and F denote the following events:
E: The card is a face card.
F: The card is a spade.
E.g. 6—The Probability of the Union of an Event
There are 12 face cards and 13 spades in a
51-card deck, so
12
P (E ) 
52
and
13
P (F ) 
52
E.g. 6—The Probability of the Union of an Event
Since three cards are simultaneously face
cards and spades, we have
3
P (E  F ) 
52
E.g. 6—The Probability of the Union of an Event
Thus, by the formula for the probability of the
union of two events, we have
P (E  F )  P (E )  P (F )  P (E  F )
12 13 3



52 52 52
11

26
The Union of
Mutually Exclusive Events
Mutually Exclusive Events
Two events that have no outcome in common
are said to be mutually exclusive.
• This is illustrated in the figure.
A Mutually Exclusive Event
For example, draw a card from a deck.
• Consider the events
E:
The card is an ace.
F:
The card is a queen.
• They are mutually exclusive because a card
cannot be both an ace and a queen.
Mutually Exclusive Events
If E and F are mutually exclusive events,
then E  F contains no elements.
• Thus,
P (E  F )  0
• So,
P (E  F )  P (E )  P(F )  P(E  F )
 P (E )  P(F )
• We have proved the following formula.
Probability of the Union of Mutually Exclusive Events
If E and F are mutually exclusive events
in a sample space S, then the probability
of E or F is
P(E  F )  P(E )  P(F )
Multiple Mutually Exclusive Events
There is a natural extension of this formula
for any number of mutually exclusive events:
• If E1, E2, … , En are pairwise mutually exclusive,
then
P E1  E2  ...  En   P E1   P E2   ...  P En 
E.g. 7—The Union of Mutually Exclusive Events
A card is drawn at random from a standard
deck of 52 cards.
• What is the probability that the card is either
a seven or a face card?
• Let E and F denote the following events:
E:
The card is a seven.
F:
The card is a face card.
E.g. 7—The Union of Mutually Exclusive Events
A card cannot be both a seven and
a face card.
• Thus, the events are mutually exclusive.
E.g. 7—The Union of Mutually Exclusive Events
We want the probability of E or F.
• In other words, the probability of E  F .
E.g. 7—The Union of Mutually Exclusive Events
By the formula,
P (E  F )  P (E )  P (F )
4 12


52 52
4

13
The Intersections of
Independent Events
The Intersection of Events
We have considered the probability of events
joined by the word or.
• That is, the union of events.
• Now, we study the probability of events joined
by the word and.
– In other words, the intersection of events.
The Intersection of Independent Events
When the occurrence of one event does not
affect the probability of another event:
• We say that the events are independent.
• For instance, if a balanced coin is tossed,
the probability of showing heads on the
second toss is 1/2.
– This is regardless of the outcome of the first
toss.
– So, any two tosses of a coin are independent.
Probability of the Intersection of Independent Events
If E and F are independent events in a
sample space S, then the probability
of E and F is
P(E  F )  P(E )P(F )
E.g. 8—The Probability of Independent Events
A jar contains five red balls and four black
balls.
• A ball is drawn at random from the jar and then
replaced.
• Then, another ball is picked.
• What is the probability that both balls are red?
E.g. 8—The Probability of Independent Events
The events are independent.
• The probability that the first ball is red is 5/9.
• The probability that the second ball is red is
also 5/9.
• Thus, the probability that both balls are red is
5 5 25
 
9 9 81
 0.31
E.g. 9—The Birthday Problem
What is the probability that in a class of
35 students, at least two have the same
birthdays?
• It is reasonable to assume that the 35 birthdays
are independent.
• It can also be assumed that each day of the
365 days in a year is equally likely as a date
of birth.
– We ignore February 29.
E.g. 9—The Birthday Problem
Let E be the event that two of the students
have the same birthday.
• It is tedious to list all the possible ways in which
at least two of the students have matching
birthdays.
• So, we consider the complementary event E′.
• That is, that no two students have the same
birthday.
E.g. 9—The Birthday Problem
To find this probability, we consider the
students one at a time.
• The probability that the first student has
a birthday is 1.
• The probability that the second has a birthday
different from the first is 364/365.
• The probability that the third has a birthday
different from the first two is 363/365.
• And so on.
E.g. 9—The Birthday Problem
Thus,
364
P (E ')  1
365
 0.186
363
365
362
331
...
365
365
• So,
P (E )  1  P (E ')
 1  0.186  0.814
The Birthday Paradox
Most people are surprised that the probability
in Example 9 is so high.
• For this reason, this problem is sometimes
called the “birthday paradox”.
The Birthday Paradox
The table below gives the probability that
two people in a group
will share the same
birthday for groups
of various sizes.