Transcript PowerPoint Presentation - Unit 1 Module 1 Sets, elements

Part 3 Module 3 Probability
To introduce the classical definition of probability, here is a question that
you can probably answer, intuitively, regardless of whether or not you
have studied probability.
Suppose we roll one die.
Q: What is the probability that the result of the die roll will be a
“two”?
A: We would probably say that the probability is “one out of six,” or
something equivalent to that.
This is because we know that when we roll one die, there are six equally
likely outcomes.
Of the six equally likely outcomes, one of them is a “two.”
Probabilities are numbers
In this course, we will express probabilities as numbers, rather
than as ratios.
Instead of saying that the probability of rolling a two is “one out of
six,” we will say that the probability is
1/6
or
.167 (the decimal equivalent of 1/6, rounded)
Calculating a probability is equivalent to calculating a percent.
Terminology: random experiment
A random experiment is any process whose
outcome isn’t known in advance.
Rolling a die is an example of a random experiment.
Terminology: sample space
The sample space (S) for a random experiment is
the set of all equally-likely outcomes.
For the experiment of rolling one die, the sample
space is
S = {1, 2, 3, 4, 5, 6}
Terminology: event
An event (E) is any outcome or combination of outcomes in a random
experiment.
An event is always a subset of the sample space for the random
experiment.
In the previous example, the experiment consisted of rolling one die, and
the event (E) we were interested in was “the result of the die roll is a
two.”
E = {2}
The classical definition of probability
Using the terminology just introduced, we have the following
classical or theoretical definition of the probability of an
event E.
P(E) 


P(E) 
number of outcomes favorable to E
number of possible equally likely outcomes to the experiment
n(E)
n(S)
Simple Example
Let F be the even that the die roll is even.
F = {2,4,6}
What is our Sample Space?
S = {1,2,3,4,5,6}
Find P(F), the probability that event F will occur.
n(F) 3 1
P(F) 
 
n(S) 6 2
The probability that we will roll an even number is 0.5.
Recall that in this class we consider probabilities as fractions, or numbers rather than ratios.
So we will read this as “one half” or “0.5”, but we will not read it as a ratio “one out of two”.

Probability
In much of our work in Part 3 Modules 3 through 5, we will work with
experiments that involve selecting one or more individuals from a specified
population. The simplest situation is one in which a single individual is
being selected.
At the Forest Folks Gathering there are
5 elves (E)
2 hobbits (H)
8 gnomes (G)
They will randomly select one person from amongst this group to serve as
Grand Marshall.
1. What is the probability that the randomly selected person is a gnome (G)?
2. What is the probability that the randomly selected person is an elf (E)?
Answers
P(G) = 8/15 = .5333
P(E) = 5/15 = .3333
Probability
At the Forest Folks Gathering there are
5 elves (E)
2 hobbits (H)
8 gnomes (G)
They will randomly select one person from amongst this group to serve as Grand
Marshall.
1. What is the probability that the randomly selected person is not a gnome (G´)?
2. What is the probability that the randomly selected person is not an elf (E´)?
Answers
P(G´) = 7/15 = .4667
P(E´) = 10/15 = .6667
Complements
Let E be any event. The complement of E, denoted
E´
is the non-occurrence of E, or the opposite of E.
In the previous examples, for instance, note that the probability of selecting
a gnome was .5333 [that is, P(G) = .5333] and the probability of not
selecting a gnome was .4667 [that is, P(G´) = .4667]
Also note that these two probabilities have a special relationship:
P(G) + P(G´) = .5333 + .4667 = 1
This illustrates an important fact called the Complements Rule.
The Complements Rule
The Complements Rule
For any event E in any experiment,
P(E) + P(E´) = 1
This rule is usually presented in a slightly different, but equivalent, form:
The Complements Rule
For any event E in any experiment,
P(E´) = 1 – P(E)
A different idea
Among a certain group of Vikings, 28 of them like to pillage, 18 of
them like to plunder, while 10 of them like to pillage and like to
plunder and 12 of them don't like to pillage and don't like to
plunder. If one of these Vikings is randomly selected, find the
probability that he/she likes to pillage.
A. 0.4828
B. 0.3103
C. 0.5833
D. 0.2800
Solution
Among a certain group of Vikings, 28 of them like to pillage, 18 of
them like to plunder, while 10 of them like to pillage and like to
plunder and 12 of them don't like to pillage and don't like to
plunder. If one of these Vikings is randomly selected, find the
probability that he/she likes to pillage. A. 0.4828
B.
0.3103
C. 0.5833
D. 0.2800
Because these categories (pillage, plunder) are
not mutually exclusive, a Venn diagram
will be helpful. The diagram shows
that the total population is 48, and 28
of them like to pillage, so the probability
is 28/48 = .5833
Another exercise
Among a certain group of Vikings, 28 of them like to pillage, 18 of
them like to plunder, while 10 of them like to pillage and like to
plunder and 12 of them don't like to pillage and don't like to
plunder. If one of these Vikings is randomly selected, find the
probability that he/she doesn’t like to pillage.
Solution: (We will refer to the answer to the previous question, and use
the complements rule)
P(doesn’t like to pillage) = 1 – P(likes to pillage)
= 1 – .5833
= .4167
Mutually exclusive events
Two events are said to be mutually exclusive if it is
impossible for the two events to occur simultaneously.
If E, F are mutually exclusive events, then
P(E or F) = P(E) + P(F)
Mutually exclusive events
At the Forest Folks Gathering there are
5 elves (E)
2 hobbits (H)
8 gnomes (G)
They will randomly select one person from amongst this group to serve as Grand
Marshall.
What is the probability that the randomly selected person is a gnome or an
elf?
Note that these events (G, E) are mutually exclusive: it is possible that the selected
person might be a gnome, and it is possible that the selected person might be
an elf, but it is impossible for the selected person to be both a gnome and an
elf.
P(G or E) = P(G) + P(E) = .5333 + .3333 = .8666
Statistical Probability
Certain statistics are also probabilities.
In this course, we will work frequently with population
statistics.
Statistical probability is inferred from population statistics.
Statistical Probability
A few years ago, for example, the Natural Resources Defense
Council conducted a study of bottled water (this example is
non-fiction).
They found that 40% of bottled water samples were merely
tap water.
They also found that 30% of bottled water samples were
contaminated by substances such as arsenic and fecal
bacteria.
Statistical Probability
40% of bottled water samples are merely tap water.
30% of bottled water samples are contaminated by substances
such as arsenic and fecal bacteria.
These unappetizing statistics are probabilities.
Let T be the event that a randomly selected sample of bottled
water is tap water.
Let C be the event that a randomly selected sample of bottled
water is contaminated.
Then
P(T) = 40% = .4
P(C) = 30% = .3
The Complements Rule, again
40% of bottled water samples are merely tap water.
30% of bottled water samples are contaminated by substances such as
arsenic and fecal bacteria.
Let T be the event that a randomly selected sample of bottled water is tap
water.
Let C be the event that a randomly selected sample of bottled water is
contaminated.
1. What is the probability that a randomly selected sample of bottled water
is not merely tap water?
2. What is the probability that a randomly selected sample of bottled water
is not contaminated?
Answers:
1. P(T´) = 1 – P(T) = 1 – .4 = .6
2. P(C´) = 1 – P(C) = 1 – .3 = .7
Mutually exclusive events
The table below shows the distribution of scores on Test 1 in Partial
Differential Equations for Liberal Arts.
Score percent
0 - 49
2%
50 - 59
6%
60 - 64
6%
65 - 69 14%
70 - 79 37%
80 - 89 17%
90 -100 18%
If one student is randomly selected, what is the probability that his/her test
score is in the 50-59 range or the 70-79 range?
A. .43 B. .97 C. .63 D. None of these
Solution
The table below shows the distribution of scores on Test 1 in Partial Differential Equations for Liberal Arts.
Score
0 - 49
50 - 59
60 - 64
65 - 69
70 - 79
80 - 89
90 -100
percent
2%
6%
6%
14%
37%
17%
18%
If one student is randomly selected, what is the probability that his/her test score is in the 50-59 range or
the 70-79 range?
These are mutually exclusive events. The probability that a student’s score in in the 50-59 range is 6%
or .06; the probability that a student’s score is in the 70-79 range is 37% or .37;
so, the probability that a student’s score is in the 50-59 range or the 70-79 range is
.06 + .37 = .43
Side note: “Odds”
Odds, like probability, use ratios or fractions to indicate likelihood.
Odds are not the same as probability, however.
According to the classical definition, that the probability of an event is
the ratio of the number of favorable outcomes to the total number of
possible outcomes.
The odds in favor of an event is the ratio of favorable outcomes to
unfavorable outcomes.
The odds against an event is the ratio of unfavorable outcomes to
favorable outcomes.
Odds vs. probability
The odds in favor of an event is the ratio of favorable outcomes to
unfavorable outcomes.
Refer to the experiment of rolling one die.
Let E be the event that the result of the die roll is “2.”
We know that P(E) = 1/6.
The odds in favor of E are not 1/6, however.
In this experiment, one of the six outcomes is favorable to E, and the other
five outcomes are unfavorable.
The odds in favor of E are 1/5, or “1 to 5.”
This is also expressed as 1:5.
The odds against E are “5 to 1”, or 5:1.
Exercise
The table below shows the distribution of scores on Test 1 in Partial
Differential Equations for Liberal Arts.
Score percent
0 - 49
2%
50 - 59
6%
60 - 64
6%
65 - 69 14%
70 - 79 37%
80 - 89 17%
90 -100 18%
If one student is randomly selected, find the ODDS AGAINST the event
that he or she has a test score in the 70 - 79 range?
A. 63:37
B. 37:100
C. 37:63
D. 63:10
Solution
Score
percent
70 - 79
37%
If one student is randomly selected, find the ODDS AGAINST the event
that he or she has a test score in the 70 - 79 range?
The odds against an event are the ratio of unfavorable outcomes to
favorable outcomes.
Since 37% of the students have scores in the 70-79 range, we know that
37(%) are favorable to that event, and 63(%) are unfavorable, so the
ratio of unfavorable to favorable outcomes is 63:37 (choice A).
A. 63:37
Odds in Sports Betting
One last note:
When the term “odds” is used in the context of sports betting, it refers to the odds
against an event.
For instance, if we are told that the odds for a certain horse winning a race are 7 to
1, that means that the odds against the horse winning are 7 to 1; the odds in
favor are 1 to 7.