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Section 4-1
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Rare Event Rule for Inferential Statistics:
If, under a given assumption, the probability of a
particular observed event is extremely small, we
conclude that the assumption is probably not
correct. * LESS THAN .05
Statisticians use the rare event rule for
inferential statistics.
Section 4-2
Basic Concepts of
Probability
Key Concept
This section presents three approaches to
finding the probability of an event.
The most important objective of this section is
to learn how to interpret probability values.
Part 1
Basics of Probability
Events and Sample Space
Event: any collection of results or outcomes of a
procedure.
Simple Event: an outcome or an event that cannot be
further broken down into simpler components.
Sample Space: for a procedure consists of all possible
simple events; that is, the sample space consists of all
outcomes that cannot be broken down any further.
Example 1: In the following display, we use “f” to denote a
female baby and “m”
to denote a male baby.
Procedure Example of Event
Complete Sample Space
Single birth
1 female (simple event)
{f, m}
3 births
2 females and 1 male
(ffm, fmf, mff, are all
simple events resulting in
2 females and a male)
{fff, ffm, fmf, fmm,
mff, mfm, mmf,
mmm}
Example 1: In the following display, we use “f” to denote a female baby and “m”
to denote a male baby.
Procedure
Example of Event
Complete Sample Space
Single birth
1 female (simple event)
{f, m}
3 births
2 females and 1 male
(ffm, fmf, mff, are all
simple events resulting in
2 females and a male)
{fff, ffm, fmf, fmm, mff,
mfm, mmf, mmm}
With one birth, the result of 1 female is a simple event because it cannot be broken
down any further. With three births, the event “2 females and 1 male” is not a
simple event because it can be broken down into simpler events, such as ffm, fmf,
or mff. With three births, the sample space consists of the 8 simple events listed
above. With three births, the outcome if ffm is considered a simple event, it is an
outcome that cannot be broken down any further. We might incorrectly that ffm
can be broken down into the individual results of f, f, and m, but f, f, and m are
not individual outcomes from three births. With three births there are exactly 8
outcomes that are simple events: fff, ffm, fmf, fmm, mff, mfm, mmf, and mmm.
Notation for Probabilities
P - denotes a probability.
A, B, and C - denote specific events.
P(A) - denotes the probability of event A
occurring.
Basic Rules for Computing Probability
Rule 1: Relative Frequency Approximation of
Probability
Conduct (or observe) a procedure, and count the
number of times event A actually occurs. Based on
these actual results, P(A) is approximated as follows:
number of times A occurred
P ( A) 
number of times procedure was repeated
Example 2: For a recent year there were 6,511,100
cars that crashed among the 135,670,000 cars
registered in the United States (based on data from
Statistical Abstract of the United States). Find the
probability that a randomly selected car in the United
States will be in a crash this year.
6,511,100
 0.048
135, 670, 000
Basic Rules for Computing Probability
Rule 2: Classical Approach to Probability (Requires Equally
Likely Outcomes)
Assume that a given procedure has n different simple events
and that each of those simple events has an equal chance of
occurring. If event A can occur in s of these n ways, then:
s
number of ways A can occur
P  A  
n number of different simple events
Example 3: There are 8 equally likely outcomes when a
woman has three babies. What is the probability that
she has 3 girls?
The possible outcomes are:
GGG
BBB
GGB
BBG
GBB
BGG
GBG
BGB
Only one of these possibilities have all three girls, so:
1
 0.125
8
Basic Rules for Computing Probability
Rule 3: Subjective Probabilities
P(A), the probability of event A, is estimated by
using knowledge of the relevant circumstances.
Law of Large Numbers
As a procedure is repeated again and again, the
relative frequency probability of an event tends to
approach the actual probability.
Probability Limits
Always express a probability as a fraction or decimal
number between 0 and 1.
The probability of an impossible event is 0.
The probability of an event that is certain to occur is 1.
For any event A, the probability of A is between 0 and
1 inclusive. That is, 0  P(A)  1.
Possible Values
for Probabilities
Complementary Events
The complement of event A, denoted by A,
consists of all outcomes in which the event A
does not occur.
Rounding Off Probabilities
When expressing the value of a probability, either give
the exact fraction or decimal or round off final decimal
results to three significant digits. (Suggestion: When a
probability is not a simple fraction such as 2/3 or 5/9,
express it as a decimal so that the number can be
better understood.)
Example 4: Some trees in a forest were showing signs of disease. A
random sample of 200 trees of various sizes was examined yielding the
following results:
a) What is the probability that one tree selected at random is large?
There are a total of 200 trees and 68 of them are large, so:
68
 0.34
200
b) What is the probability that one tree selected at random is diseased?
There are a total of 200 trees and 37 of them are diseased, so:
37
 0.185
200
Example 4 continued: Some trees in a forest were showing signs of
disease. A random sample of 200 trees of various sizes was examined
yielding the following results:
c) What is the probability that one tree selected at random is both small
and diseased?
There are a total of 200 trees and 8 of them are small AND diseased, so:
8
 0.04
200
d) What is the probability that one tree selected at random is both
doubtful and medium?
There are a total of 200 trees and 32 of them are medium AND
32
doubtful, so:
 0.16
200
Example 5: Express the indicated degree of likelihood
as a probability value between 0 and 1 inclusive. In
one of New York State’s instant lottery games, the
chance of a win are stated as “4 in 21”.
4
 0.19
21
Example 6: Express the indicated degree of
likelihood as a probability value between 0 and
1 inclusive. If you make a random guess for the
answer to a true/false test question, there is a
50-50 chance of being correct.
1
 0.50
2
Example 7: Express the indicated degree of
likelihood as a probability value between 0 and
1 inclusive. It is impossible to get five aces
when selecting cards from a shuffled deck.
0
0
52
Example 8: A couple decides to have three children. Find the probability of
each event.
a) There is exactly one boy.
Let’s recall our sample space: GGG
BBB
GGB
BBG
GBB
BGG
GBG
BGB
Of those 8 possibilities, 3 of them have exactly one boy, so:
3
 0.375
8
b) There are exactly two boys.
Of the 8 possibilities from above, 3 of them have exactly two boys, so: 3  0.375
8
c) All are boys.
Of the 8 possibilities from above, 1 of them have all boys, so: 1  0.125
8
Example 9: The 110th Congress of the United States
included 84 male Senators and 16 female Senators. If
one of these Senators is selected, what is the
probability that a woman is selected? Does this
probability agree with a claim that men and women
have the same chance of being elected as Senators?
There are 84 + 16 = 100 total senators.
Let W = selecting a woman. P(W) = 16/100 = 0.16.
No, this probability is too far below 0.50 to agree with
the claim that men and women have equal
opportunities to become a senator.
Example 10: In the last 30 years, death sentence executions in
the United States included 795 men and 10 women (based on
data from the Associated Press). If an execution is randomly
selected, find the probability that the person executed is a
woman. Is it unusual for a woman to be executed? How might
this discrepancy be explained?
There were 795 + 10 = 805 total persons executed.
Let W = a randomly selected execution was that of a woman.
P(W) = 10/805 = 0.0124.
Yes. Since 0.0124  0.05, it is unusual for an executed person to
be a woman. This is due to the fact that more crimes worthy of
the death penalty are committed by men than women.
“Rare event rule for inferential statistics” from Section 4.1.
Example 11: Each of two parents has the genotype brown/blue,
which consists of the pair of alleles that determine eye color, and
each parent contributes one of those alleles to a child. Assume
that if the child has at least one brown allele, that color will
dominate and the eyes will be brown. (The actual determination
of eye color is somewhat more complicated.)
a) List the different possible outcomes. Assume that those
outcomes are equally likely.
Listed with the father’s contribution first, the sample space has 4
simple events: brown/brown brown/blue blue/brown
blue/blue
b) What is the probability that a child of these parents will have
blue/blue genotype?
P(blue/blue) = ¼ = 0.25
Example 11 continued:
c) What is the probability that a child will have brown
eyes?
P(brown eyes) = P(brown/brown or brown/blue or
blue/brown) = ¾ = 0.75
Part 2:
Beyond the Basics of Probability-Odds
Odds
The actual odds against event A occurring are the ratio P(A)/P(A),
usually expressed in the form of a:b (or “a to b”), where a and b
are integers having no common factors.
The actual odds in favor of event A occurring are the ratio
P(A)/P(A), which is the reciprocal of the actual odds against the
event. If the odds against A are a:b, then the odds in favor of A
are b:a.
The payoff odds against event A occurring are the ratio of the net
profit (if you win) to the amount bet.
payoff odds against event A = (net profit) : (amount bet)
Example 12: When the horse Barbaro won the 132nd Kentucky
Derby, a $2 bet that Barbaro would win resulted in a return of
$14.20.
a) How much net profit was made from a $2 win bet in Barbaro?
If a $2 bet results in a net return of $14.20, the net profit is 14.20
– 2.00 = $12.20.
b) What were the payoff odds against a Barbaro win?
If you get back your bet plus $12.20 for every $2 you bet, the
payoff odds are 12.2 : 2 (We express this as 6.1 : 1 or 61 : 10), or
approximately 6 : 1.
c) Based on preliminary wagering before the race, bettors
collectively believed that Barbaro had a 57/500 probability of
winning. Assuming that 57/500 was a true probability of s
Barbaro victory, what were the actual odds against his winning?
Odds against W = P(not win)/P(W) = (443/500)/(57/500) = 443/57
or 443 : 57. About 8 : 1.
d) If the payoff odds were the actual odds found in part (c), how
much would a $2 win ticket be worth after the Barbaro win?
If the payoff odds are 7.77 : 1, a win gets back your bet plus $7.77
for every $1 bet. If you bet $2 and win, you get back 2 + $17.54.
The ticket would be worth $17.54 and your profit would be 17.54
– 2.00 = $15.54.