Section 11.3Basic Concepts of Probability

Math in Our World

Learning Objectives

Compute classical probabilities. Compute empirical probabilities.

Probability Experiments

A probability experiment is a process that leads to well-defined results called outcomes.An outcome is the result of a single trial of a probability experiment.

Example: flipping a coin and getting heads

Probability Experiments

A sample space is the set of all possible outcomes of a probability experiment.

Experiment Sample Space

Toss one coin {head, tail}

Roll a die {1, 2, 3, 4, 5, 6}

Toss two coins {head/head, tail/tail, head/tail, tail/head}

Probability Experiments

An event is any subset of the sample space of a probability experiment.

Experiment: draw a card.Sample space: 52 cards.Event: drawing an ace.

Classical Probability

Let E be an event that is a subset of the sample space S. We will write n(E) to represent the number of outcomes in E, and n(S) to represent the number of outcomes in S. The probability of the event E is defined to be

There is one key assumption we make in classical probability: that every outcome in a sample space is equally likely.

( ) Number of outcomes in ( )

( ) Number of outcomes in

n E EP E

n S S= =

EXAMPLE 1 Computing Classical Probabilities

A single die is rolled. Find the probability of getting

(a) A 2.

(b) A number less than 5.

(c) An odd number.

In this case, since the sample space is 1, 2, 3, 4, 5, and 6, there are six outcomes: n(S) = 6.(a) There is one possible outcome that gives a 2, so

SOLUTION

1(2)

6P =

EXAMPLE 1 Computing Classical Probabilities

(b) There are four possible outcomes for the event of getting a number less than 5—1, 2, 3, or 4; so n(E) = 4, and

SOLUTION

(c) There are three possible outcomes for the event of getting an odd number: 1, 3, or 5; so n(E) = 3, and

( )(a number less than 5)

( )

n EP

n S=

4 2

6 3= =

( )(odd number)

( )

n EP

n S= 3 1

6 2= =

EXAMPLE 2 Computing Classical Probabilities

Two coins are tossed. Find the probability of getting

(a) Two heads.

(b) At least one head.

(c) At most one head.

SOLUTIONThe sample space is {HH, HT, TH, TT}; therefore, n(S) = 4.(a) There is only one way to get two heads: HH. So

( )(two heads)

( )

n EP

n S= 1

4=

EXAMPLE 2 Computing Classical Probabilities

(b) “At least one head” means one or more heads; i.e., one head or two heads. There are three ways to get at least one head: HT, TH, and HH. So n(E) = 3, and

SOLUTION

(c) “At most one head” means no heads or one head: TT, TH, HT. So n(E) = 3, and

( )(at least one head)

( )

n EP

n S=

3

4=

( )(at most one head)

( )

n EP

n S= 3

4=

Facts About Probability

• Both n(E) and n(S) have to be zero or positive, so probability is never negative.

• Since S represents all possible outcomes, and E is a subset of S, n(E) is always less than or equal to n(S). This means that probability is never greater than 1.

• When an event cannot possibly occur, the probability is zero, and when an event is certain to occur, the probability is one.

Facts About Probability

• The sum of the probabilities of all the outcomes in the sample space will always be one.

• For any event, E, P(E') = 1 − P(E), where E' is the event “E does not occur.”For example, if a die is rolled, the probability that a 4 will not occur, symbolized by P(4'), is 5/6 since there are five ways that a 4 will not occur; i.e., 1, 2, 3, 5, 6. OR

Empirical Probability

Empirical probability is based on observed frequencies—that is, the number of times a particular event has occurred out of a certain number of trials.

EXAMPLE 3 Computing an Empirical Probability

In a sample of 50 people, 21 had type O blood, 22 had type A blood, 5 had type B blood, and 2 had type AB blood. Set up a frequency distribution and find these probabilities for a person selected at random from the sample.(a) The person has type O blood.

(b) The person has type A or type B blood.

(c) The person has neither type A nor type O blood.

(d) The person does not have type AB blood.Source: Based on American Red Cross figures

EXAMPLE 3 Computing an Empirical Probability

SOLUTION

(a)

(b) The frequency of A or B is 22 + 5 = 27.

Type (class) Observed Frequency

A 22

B 5

AB 2

O 21

Total 50

21(O)

50

fP

n= =

27(A or B)

50P =

(c) Neither A nor O means either B or AB; the frequency of either B or AB is 5 + 2 = 7. 7

(neither A nor O)50

P =

(d) The probability of not AB is found by subtracting the probability of AB from 1.

(not AB) 1 (AB)P P= −2 48 24

150 50 25

= − = =