STT200
Chapter 14-15
K.A. M
Chapter 14 From Randomness to Probability
Probability theory helps to measure a “likelihood” of an event, and answer the
questions like:
 How likely is it to answer correctly one out of two true-false questions on a quiz?
 Is it more, less, or equally likely as guessing five out of ten true-false questions?
 Is it likely or unlikely to answer all ten true-false questions on a quiz if all answers
are simply guessed?
 If you roll two dice, is it more likely to roll the sum of ten, or the sum of eight?
 What does it mean “There is thirty percent chance for a rain tomorrow”?
You will be able to answer the questions above (and more) with confidence when we
finish this and next chapter.
Consider the following experiment:
Adam has taken a 10 True-False questions quiz. He claims that he came totally
unprepared and had to select each answer by tossing a coin. And he got a B! Should
we believe it? Why or why not? (remember the question while studying binomial
distribution).
Rare Event Rule for Inferential Statistics: If under given assumptions (like truly
random guessing) the probability of an observed event (like guessing randomly, and
getting all 10 answers correct) is very small, then we should conclude that the
assumption was wrong.
Probability deals with modeling of random phenomena (phenomena, or experiments
whose outcomes may vary)
Each observation of a random experiment is called a trial.
The value of the random experiment is called an outcome.
When we combine some outcomes, the resulting combination is an event.
The collection of all possible outcomes is called the sample space.
Examples of random experiments:
1. A toss of a coin
2. A roll of a die
3. A random selection of an individual from a population
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Intuitively probability of an event is a long-run proportion of times the event occurs in
many independent repetitions of the experiment.
Law of Large Numbers: the long-run relative frequency of repeated event gets closer
and closer to a single value called the probability of the event.
This definition of probability is often called empirical probability
# of times A has occured
 P( A)
# of repetitions
Experiment : Toss a coin 10 times. Write down the proportion of heads.
Case #
Proportion of
heads
After 1 toss
After 2 tosses
After 3 tosses
…
(we can simulate tosses using the calculator: MATH-PROB-5:RandInt(0,1)
Enter)
Sketch a scatterplot: proportion of heads vs. number of tosses
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The more tosses, the more sure we are that the coin falls head up …. % of the
time. This coin seems to be …. (fair? Biased?)
How about another coin, tossed 100 times, represented by the graph below?
Proportion and probability. The following questions are equivalent:
 What proportion of all such families has two girls?
 How likely is it to select randomly a family with two girls?
 What is the probability that a randomly selected family with two kids has only
girls?
Example: In a given very large batch of calculators 5% are defective. Therefore, the
probability of selecting at random a defective calculator is 5%.
Here is a nice link which might help you understand the frequency distribution of
the outcomes using the example of throwing a die (or two) many times:
http://gwydir.demon.co.uk/jo/probability/dice.htm
It is often HARD to find the probability of a given event! But we can approximate it quite
well by simulating the event.
Have you heard of the Three Doors game? Check below and try yourself.
http://math.ucsd.edu/~crypto/Monty/monty.html
Model of a random experiment:
1. The sample space S = the set of all possible outcomes
2. Events = subsets of the sample space - illustrated on Venn Diagrams (the rectangle
represents the entire sample space S, and the circles inside represent events of
interest)
3.
Probability = a function that assigns a number to each event A. That number is
denoted as P(A) (called the probability of an event A)
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Illustration of events – Venn diagram
A
If A is an event, the event that "A does not occur" is denoted Ac and is called the
complement of A
A
AC
Two events are called disjoint, if they cannot occur simultaneously
A
B
Combined events: A and B, A or B
A and B
A or B
Union of A and B Union of A and
B
Probability is denoted as P(A) (called the probability of an event A) with the following
properties:
 0  P(A)  1
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 P(S) = 1 (S is sample space and represents the set of all possible outcomes.)
 Events A and AC are complementary events if they are disjoint (mutually
exclusive, no common outcomes), and P(Ac) = 1- P(A)
Simplified addition and multiplication rules:
 if A are B are disjoint then
P(A or B) = P(A) + P(B)
Events A and B are independent if P(A and B) = P(A)·P(B)
Example: sample space, outcomes, events, probabilities
A coin is tossed three times.
a) Find S (sample space) of this experiment
Solutions: The sample space
S = {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT}
b) Let A = "two heads (H) in three tosses"
B = "H in first toss"
C = "H in second toss"
Find the probability of the events A, B, C.
Solutions: First find the list of simple events from sample space matching each
description:
A= {HHT, HTH, THH}
B= {HHH, HHT, HTH, HTT }
C= {HHH, HHT, THH, THT }
Probability of each outcome = 1/8: equally likely outcomes with the total of 8 outcomes
P(A) = 3/8
P(B) = 4/8 = 1/2
P(C) = 4/8 = 1/2
More examples of combined events:
E={B and C} = “ H in first toss and H in second toss” = {HHH, HHT}
F={B or C} = “ H in first toss or H in second toss” =
= { HHH, HHT, HTH, HTT, THH, THT }
Independence of the events
Are A and B above independent?
QUESTION: How to check if two events A and B are independent?
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ANSWER: Show that P(A and B)=P(A)·P(B)
The event {A and B} =" two heads in three tosses and H in the first toss} H" = {HHH,
HHT}
P(A and B) = 2/8 =1/4
P(A)=3/8
P(B) =2/8
P(A and B) =? =P(A)· P(B)
What Can Go Wrong?

Beware of probability assignments where probabilities that don’t add up to 1.
To be a legitimate probability assignment, the sum of the probabilities for all possible
outcomes must total 1.

Don’t “just add” probabilities of events if they’re not disjoint. Events must be
disjoint to use the Basic Addition Rule.

Don’t “just multiply” probabilities of events if they’re not independent.
Direct multiplication of probabilities of events that are not independent is one of the
most common errors people make in dealing with probabilities.

Don’t confuse disjoint and independent—disjoint events can’t be independent
(why?)
Class exercises:
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STT 200 Classwork
Chapter 14-15
K.A. M
Chapter 14
A white and black dice are tossed
1. Identify the events:
 A ="sum equals 6",
 B = "sum is at most 5" ,
 C= "same number on each
die",
 D ="at least one shows 4"
and find their probabilities
n(A)=
n(B)=
n(C)=
n(D)=
P(A)=
P(B)=
P(C)=
P(D)=
2. Find the probability of A or B.
Are the events A and B
independent? Are they disjoint?
3. Let W="white shows 4" and
V="black shows 4"
Are they disjoint?
Are the events W and V
independent?
(Hint: if they are independent, then the equation P(A and B) = P(A)·P(B) holds true)
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Classwork
Chapter 14 B
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What had to be assumed to answer last four questions?
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Chapter 15
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Probability Rules!
Brief review
Model of a random occurrence (discussed in Chapter 14)
1. Sample space S = the set of all possible outcomes
2. Events = subsets of the sample space
 Ac = "A does not occurs" - the complement of A
 Two events are called disjoint, if they cannot occur simultaneously
 Combined events: A or B, A and B
3. Probability = a function that to each event A assigns a number P(A)
 0  P(A)  1
 P(S) = 1
 P(Ac) = 1-P(A)
 if A are B are disjoint, then P(A or B) = P(A) + P(B)
 If A are B are independent, then P(A and B) = P(A) P(B)
Model with Equally Likely Outcomes: If the sample space has finitely many possible
outcomes and all outcomes are equally likely, then the probability of an event A is
P( A) 
count of outcomes in A
count of all possible outcomes
Many models do not have this property.
General addition rule: if two events are NOT disjoint then
P(A or B) = P(A) + P(B) - P(A and B)
if two events are disjoint then P(A and B)=0 and the “basic” addition rule works.
Examples:
1.
Let’s count how many people in our classroom have dogs. How many have cats.
How many have both.
Find the probability that a randomly selected person in our class has a dog or a cat.
Are the two events: “having a dog” and “having a cat” disjoint? Are they independent?
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2. Select a single card from a deck.
What is the probability of drawing a king or a diamond? ____________
Example. Police report that 78% of drivers stopped on suspicion of drunk driving are
given a breath test, 36% a blood test, and 22% both tests. What is the probability that a
randomly selected DUI suspect is given
1. At least one test?
2. A blood test or a breath test, but not both?
3. Neither test?
0.08
0.14
0.56
0.22
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Ans: 0.78 + 0.36 – 0.22 = 0.92
Ans: 0.70 (see diagram)
Ans: 0.08 (see diagram)
One more concept: Conditional Probability
Gender
Example. Given is a contingency table of students cross-classified by their school
goal and gender
Boy
Girl
Total
Grades
24
27
51
Goals
Popular
Sports
10
13
19
7
29
20
Total
47
53
100
A student is selected at random. Let
G ="a girl is selected" and S = "wants to excel at sports"
1. Find P(G)
Ans: 53/100 = 0.53
2. Find P(S)
Ans: 20/100 = 0.20
3. Find the probability that "a girl is selected and she wants to excel at sports"
P(G and S) = 7/100 = 0.07
4. Find the probability that "a student wants to excel at sports, given that a girl is
selected ".
P(S | G)
# ( S and G )
= 7/53 = 0.13
# (G )
Conditional probability of B given that A has occurred
is defined by
P( B | A) 
P( A and B)
P( A)
Conditional probability of A given that B has occurred
is defined by
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P( A | B) 
K.A. M
P( A and B)
P( A | B) means P( A if B has alreadyoccured)
P( B)
General multiplication rule
From
P( B | A) 
P( A and B)
it follows that
P( A)
P( A and B)  P( A) P( B | A)
or
P( A and B)  P( B) P( A | B)
Recall that events A and B are independent, if P(A and B) = P(A) P(B).
Thus, A and B are independent means:
P( B | A)  P( B)
Example. In a lot of 10 elements there are two defectives. Two elements are
selected at random and without replacement, one after another.
1. What is the probability that the first is good and the second is defective?
2. What is the probability that both are defective?
3. What is the probability that the second is defective?
Tree Diagram A tree diagram helps us think through conditional probabilities by
showing sequences of events as paths that look like branches of a tree.
8/10
7/9
G2
2/9
D2
8/9
G2
G1
●
2/10
D1
1/9
D2
Answers:
Let G=”selecting good one” and D=”selecting a defective one”
1. P(G1 and D2)= (8/10) * (2/9) = 8/45
2. P(D1 and D2)= (2/10)*(1/9)=1/45
P(G1 and D2) + P(D1 and D2) = (8/10)*(2/9) + (2/10)*(1/9) = 8/45+1/45=9/45=0.2 or
20%
Reversing Conditional Probability (Bayes’ Rule)
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Suppose we want to know P(A|B), but we know only P(A), P(B), and P(B|A).
We can compute P(A and B) from the formula P(A and B) = P(A)·P(B|A)=
P(B)·P(A|B)=
From this information, we can find P(A|B):
P(A|B)  P(A and B)
P(B)
(or, by re-labeling, find P(B|A))
Example: Reversing the Conditional Probability
Example. In April 2003 Science magazine reported a new test for ovarian cancer.
Ovarian cancer afflicts 1 out of 5000 women. The test is highly sensitive, correctly
detecting 99.97% of women who have the disease. But it’s unlikely that it will be
used because it gives false positives 5% of the time. Why is it a big problem? Draw
a probability tree and determine the probability that a woman who tests positive
actually has ovarian cancer.
0.9997
+
C
0.0002
0.0003
-
0.05
+
0.95
-
●
0.9998
H
What have we learned?
 The probability rules from Chapter 14 only work in special cases—when
events are disjoint or independent.
 We now know the General Addition Rule and General Multiplication Rule.
 We also know about conditional probabilities and that reversing the
conditioning can give surprising results.
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 Venn diagrams, tables, and tree diagrams help illustrate problems involving
computing probabilities and organize our thinking about probabilities.
 We now know more about independence—a sound understanding of
independence will be important throughout the rest of this course.
Class exercises Chapter 15
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BONUS up to +3 (due next class): A company manufacturing electronic components for home
entertainment systems buys electrical connectors from three suppliers. The company prefers to use supplier A
because only 2% of those connectors prove to be defective, but supplier A can deliver only 75% of the
connectors needed. The company also must purchase connectors from two other suppliers, 15% from
supplier B and the rest from supplier C. The rates of defective connectors from B and C are 3% and 5%,
respectively. You buy one of these components, and when you try to use it you find that connector is
defective.
Make a probability tree to illustrate the problem. What is the probability that your component came from
supplier A?
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