Presentation 5.
Probability
What are we doing now?

So far in this course we have covered:
1. Descriptive Statistics (Plots and Summaries of
Variables)
2. Gathering Data (types of studies and design issues)
3. Introduction to Inference (statistical procedures to
answer questions about the population; so far we have
seen the chi-square test.)
Now we are going to learn about probability and random
variables which will help us understand inference more
completely (i.e. where p-values come from, how are we able
to make conclusions based on samples…etc.)
Random Circumstances

A random circumstance is one in which the
outcome is unpredictable.
Examples:
1. Selecting a card at random from a deck.
2. Rolling a six sided die.
3. Selecting an individual at random and
recording their eye color.
4. Selecting an individual at random and
recording their weight.
Probability

Probability is how likely a particular outcome will
be the result of a random circumstance.
Examples:
1. Probability
2. Probability
3. Probability
4. Probability
lbs.
of selecting a spade.
of rolling a 4.
the person you selected has green eyes.
the person you select weighs more than 200
More on Probability
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Probability is a number between 0 and 1.
The sum of the probabilities for all outcomes of a
random circumstance should add up to 1 (or
100%).
The set of all possible outcomes of a random
circumstance is called sample space. For
example:
1.
2.
3.
Tossing a coin: sample space = {head, tail}
Rolling a die: sample space = {1,2,3,4,5,6}
Rolling two dice: sample space = {(1,1),(1,2),…,(6,6)}
Relative Frequency Interpretation
of Probability
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The word probability has two different
interpretations. One is the statistical definition,
the other (which might be slightly different) is
the use of the word probability in every day
life.
In situations that we can image repeating
many times we define the probability of a
specific outcome as the proportion of times it
would occur over the long run.
This is also called the relative frequency of that
particular outcome.
Example
0.6
0.5
0.3
0.4
Proportion of Heads
0.7
0.6
0.5
0.4
Proportion of Heads
0.8
Consider Tosses of a Coin, Let us Plot the
Proportion of Heads on the n th Toss
0
200
400
600
Number of Flips
800
1000
0
100
200
300
Number of Flips
400
500
Example
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Suppose we toss a fair coin 2,000 times. It is almost
IMPOSSIBLE for us to get exactly 1,000 heads and
1,000 tails. However, the proportion of the heads and
tails as we make more and more tosses will approach
0.5 (50%).
We have to consider the outcome in the long run.
Suppose we toss a coin 3 times and get 1 head and 2
tails. Based on this information, we cannot assign 1/3
as the probability of getting a head, and 2/3 as the
probability of getting a tail. Lack of repetition creates
a problem in this situation. To correct this error, we
need to toss the coin a large number of times.
Determining the Relative Frequency
Probability of an outcome
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Method 1: Make an Assumption about the
Physical World.
Example 7.2 A Simple Lottery
Choose a three-digit number between 000 and 999. Player wins if his or her
three-digit number is chosen. Suppose the 1000 possible 3-digit numbers
(000, 001, 002, . . . , 999) are equally likely. In long run, a player should win about 1
out of 1000 times. This does not mean a player will win exactly once
in every thousand plays.
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Method 2: Observe the Relative Frequency
Example 7.4 The Probability of Lost Luggage
“1 in 176 passengers on U.S. airline carriers will temporarily lose their
luggage.”
Example 1: A fair die is rolled once. Determine the probability
of each of the following outcomes
a. Six dots:
b. One or two dots:
c. An even number of dots:
d. more than 5 dots:
e. more than or equal to 5 dots:
f. less than 5 dots:
Example 2: When 190 students were asked to pick a
number from 1 to 10, the number of students selecting
each number were as follows:
Number:
Frequency:
1
2
2
9
3
22
4
21
5
18
6
23
7
56
8
19
9
14
10
6
TOTAL
190
What is the approximate probability that someone asked to
pick a number from 1 to 10 will pick
1.
The number 3?
2.
One of the two extremes, 1 or 10?
3.
Odd number?
Some Definitions
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The sample space is the collection of all possible
outcomes of a random circumstance (we will denote it
with S)
A simple event is one possible outcome of a random
circumstance.
An event is a collection of one or more simple events.
Example: Consider rolling a die once.
The sample space is S = {1,2,3,4,5,6}
Simple events in the sample space are: {1}, {2}, {3},
{4}, {5}, {6}.
Event to obtain an odd outcome, i.e. {1,3,5}, is
comprised of the simple events {1}, {3}, {5}.
Assigning Probabilities to Simple
Events
P(A) = probability of the event A
Conditions for Valid Probabilities
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Each probability is between 0 and 1.
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The sum of the probabilities over all
possible simple events is 1.
Equally Likely Simple Events
If there are k simple events in the sample space
and they are all equally likely, then the probability of
the occurrence of each one is 1/k.
Complementary Events
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The complement of the event A, denoted by Ac, is
the collection of all simple events that are not in A.
Ac
Note: P(A) + P(AC) = 1
Example 7.2 A Simple Lottery (cont)
A = player buying single ticket wins
AC = player does not win
P(A) = 1/1000 so P(AC) = 999/1000
Mutually Exclusive Events
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Two events are mutually exclusive,
or equivalently disjoint, if they do not contain
any of the same simple events (outcomes).
A
B
Example 7.2 A Simple Lottery (cont)
A = all three digits are the same.
B = the first and last digits are different
The events A and B are mutually exclusive
(disjoint), but they are not complementary.
Independent Events
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Two events are independent of each other if
knowing that one will occur (or has occurred)
does not change the probability that the other
occurs.
Are mutually exclusive events independent?
Conditional Probability
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Conditional probability of the event B, given
that the event A occurs, is written as P(B|A).
1.
If A and B are independent then P(B|A) =
2.
If A and B are mutually ex. then P(B|A) =
Rules for Finding Probabilities
Probability Axioms
1.
0≤ P(A) ≤ 1, for every event A.
2.
P(S) = 1
3.
For A1,A2,… disjoint events
P(A1 or A2 or … ) =P(A1 ) + P(A2 ) + …
Note:
A
B
The event “A or B” is the
whole grey area.
The event “A and B” is the
darkest are of the
Consequences of these axioms…
some basic rules
Rule 1 : P(AC) = 1 – P(A)
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since P (A or Ac ) = P( S ) = 1 and also,
P(A or Ac ) = P(A )+ P(Ac)
For A1, A2 disjoint: P(A1 and A2) = P(Ø) = 0, where Ø denotes
an event that cannot occur.
Rule 2 : P(A or B) = P(A) + P(B) – P(A and B)
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For A and B disjoint: P(A or B) = P(A) + P(B)
Rule 3 : P(A and B) = P(A)P(B|A)
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For A and B independent: P(A and B) = P(A)P(B)
For several independent events,
P(A1 and A2 and … and An) = P(A1)P(A2)…P(An)
Rule 4 : P(B | A) = P(A and B)/P(A)
P(A|B) = P(A and B)/P(B)
Example 7.12 Probability Two Strangers
Both Share Your Birth Month
Event A = 1st stranger shares your birth month P(A) = 1/12
Event B = 2nd stranger shares your birth month P(B) = 1/12
Note: Events A and B are independent.
P(both strangers share your birth month)
= P(A and B) = P(A)P(B) = (1/12)(1/12) = 0.007
Note: The probability that 4 unrelated strangers all share
your birth month would be (1/12)4.
Example 7.13 Alicia Answering
There 50 students, the professor randomly select students to
answer 3 questions.
If we know Alicia is picked to answer one of the
questions, what is the probability it was the first question?
A = Alicia selected to answer Question 1, P(A) = 1/50
B = Alicia is selected to answer any one of the questions,
P(B) = 3/50
Since A is a subset of B, P(A and B) = 1/50 (draw a diagram)
P(A|B) = P(A and B)/P(B) = (1/50)/(3/50) = 1/3
Example 1
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Studies on depression indicate that a particular course of
treatment improves the condition of 72% of those on
whom it is used, does not affect 10%, and worsens the
condition of the rest. A person suffering from depression
is treated by using this method. What is the probability
that his condition will worsen? What is the probability
that the treatment is not detrimental to his condition?
Example 2
Data gathered at a particular blood center show that
.1% of all donors test positive for HIV and 1% test
positive for herpes. If 1.05% test positive for one or the
other of these problems
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1.
2.
3.
What is the probability that a randomly selected donor will have
neither problem?
Would you be surprised to find a donor with both problems?
Are these two problems independent?
Example 3
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Approximately 50% of the population is male, 68% of the
population drinks to some extent, and 38.5% drinks to
some extent and is male. Given that a randomly selected
individual is male, find the probability that he drinks. Is a
person's drinking status independent of gender?
Is the same true if 34% drinks to some extent and is
male?