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STT 351 Conditional Probability. Independence 2.4 -2.5
1
2.4. Conditional Probability
Example 2.4.1. (#46 p.80 textbook)
Suppose an individual is randomly selected from the population of all adult males
living in the United States. Let A be the event that the selected individual is over
6 ft in height, and let B be the event that the selected individual is a professional
basketball player. Which do you think is larger,
6 ,
the individual being more than feet tall knowing
that the individual is a professional basketball plP
ayer
or
the probability of the individual being a professional basketball player, P
knowing that the individual is more than 6 feet tall
Answer with reasoning.
• Let A be that the individual is more than 6 feet tall.
• Let B be that the individual is a professional basketball player.
• P(A|B) = the probability of the individual being more than 6 feet tall,
knowing that the individual is a professional basketball player.
• P(B|A) = the probability of the individual being a professional basketball
player, knowing that the individual is more than 6 feet tall.
Because most professional basketball players are tall, so the probability of an
individual in that reduced sample space being more than 6 feet tall is very large.
On the other hand, the number of individuals that are pro basketball players is
small in relation to the number of males more than 6 feet tall
P(A|B) P(B|A)
Example 2.4.2.
What the probability that when rolling a die the side 6 will be on the top if it is
known that occurred an even number?
Answer. 1
3
Objectives.
• Definition of conditional probability and multiplication rule
• Total probability
• Bayes’ Theorem
STT 351 Conditional Probability. Independence 2.4 -2.5
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Example 2.4.4.
The population of a particular country consists of three ethnic groups.
Each individual belongs to one of the four major blood groups.
The accompanying joint probability table gives the proportions of individuals in
the various ethnic group–blood group combinations.
P C R and P C
Suppose that an individual is randomly selected from the population, and define
events ,A type A selected B type B selected , and
3C ethnic group selected
a) Calculate ,P A P C and P A C
b) Calculate both P A C and P C A and explain in context what each
P C R and P C probability represents.
Definition of Conditional Probability 2.4.3.
Conditional probability of an even A is the probability of A computed with
information that another event B occurred.
Notation: P A B
Formal definition.
For any two events , 0,A and B P B the conditional probability of A given
that B has occurred is defined by
P A BP A B
P B
STT 351 Conditional Probability. Independence 2.4 -2.5
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Solution.
0.106 0.141 0.200 0.447P A ;
0.215 0.200 0.065 0.020 0.500P C
0.200P A C
0.200
type selected given ethnic group 3 0.400
00.50
P A C A ;
0.200
ethnic group 3 given type 0.44 selected0
7.447
P C A A
2.4.6. Multiplication Rule.
Example 2.4.7.
Two cards are drawing one after one from a deck of 52 cards.
What is the probability that both cards will be aces?
Solution.
I is an ace II is an ace
4 3 1 1first ace and second ace
52 51 13 1 2217
1P
4 aces
48 others
2.4.5. Remark about Contingency Table.
In the problem given data is organized in so-called contingency table.
The contingency table is the method of organizing data for qualitative variables
as a table in which every cell represents one of the categories of a two-
directional classification.
In the example above sample consists of elements that have two features
(specified blood type and specified ethnic group). The table contains
proportions of elements that have different combinations of two variables,
blood type and ethnic group. For example, the number 0.082 (first column, first
row) is the proportion in the sample of people with blood type O and classified
as the ethnic group 1.
STT 351 Conditional Probability. Independence 2.4 -2.5
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2.4.8. The Law of Total Probability.
Consider a system 1, ... , kA A -
mutually exclusive and exhaustive events.
This means that mutually exclusive events
1, ... , kA A make partition of the sample space :
1 2 ... ,
, , , 1,
n i jA A A A A
for any i and j i j i j k
The probability of any other event B (related to the same sample space) can by found
by so-called the Formula for Total Probability
1 1
1
k k
k
i i
i
P B P A P B A P A P B A
P A P B A
2.4.9. Idea of Proof .
If in a sample space is present a system of mutually exclusive and
exhaustive events (hypotheses), then any event B connected with this experiment
will take place only together with one of the hypotheses.
So, “find the probability of the event B” means “find the probability of the union
of mutually exclusive events
STT 351 Conditional Probability. Independence 2.4 -2.5
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2.4.10. Probability Tree Diagram.
When applying the Formula for Total Probability, it is very helpful to construct so-
called Probability Tree Diagram.
In a Probability Tree Diagram:
• first set of branches (arrows) presents mutually exclusive events 1, ... , kA A
(often called “hypotheses”).
• from the end of each branch (arrow) in first set we draw second level of
branches (arrows) in correspondence with options {B given that occurred}iA
• along each arrow is written corresponding probability.
• number of branches at each level can be arbitrary, but the sum of all
probabilities at each level that go from the same node is always equal to 1.
Example.2.4.11.
Consider taking two balls, one after one, from the box that contains 2 blue balls and
3 red balls. Construct the probability tree for this experiment.
Probabilities of
hypotheses
, 1,2iP A i
First draw
Second draw
Conditional probabilities
i iP B A and P B A
with each hypothesis
Example of computing.
Pretend the event of interest B is
{the ball in the second draw is blue}.
1 1 2 2P B P A P B A P A P B A
2 1 3 2
5 4 5
7
204P B
Hint. We were doing multiplication along
branches that have blue ball at the end.
STT 351 Conditional Probability. Independence 2.4 -2.5
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Example 2.4.12.
40% of the students enrolled in a business statistics course had previously taken a
finite math course.
30 % of these students received 4.0 grade for the statistics course, whereas 20% of
other students received 4.0 grade for the statistics course.
a) Draw a tree diagram and label it with all appropriate probabilities.
b) What is the probability that a student selected at random
receive 4.0 in the statistics course?
To find the probability of the
event {student recieved 4 points}B ,
we need choose sequences of
arrows that end up with 4.0,
multiply probability along each
composite branch and add all such
products.
{student recieved 4.0}
0.4 0.3 0.6 0.2 0.24
P
STT 351 Conditional Probability. Independence 2.4 -2.5
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2.4.16. Bayes’ Theorem.
Example 2.4.18. (Example 2.31 p.79 textbook, modified).
Incidence of a rare disease. Only 1 in 100 adults is afflicted with a rare disease for
which a diagnostic test has been developed. The test is such that when an
individual actually has the disease, a positive result will occur 99% of the time,
whereas an individual without the disease will show a positive test result only
0.4% of the time. If a randomly selected individual is tested and the result is
positive, what is the probability that the individual has the disease?
The Formulation of Bayes’ Theorem.
Let be mutually exclusive and exhaustive events with prior
probabilities
Then for any other event
,
2.4.17. Idea of Proof.
Given, that the event B occurred, we don’t need any more to consider the entire
sample space. You will deal only with its part B.
To find jP A B for any jA , we need to count only outcomes that are common
for jA and B and divide this number by number of outcomes in B.
STT 351 Conditional Probability. Independence 2.4 -2.5
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Solution.
For solving the problem, we will use the Bayes’ formula.
1) Make a Probability Tree Diagram.
2) In the denominator of Bayes’ formula must be written test is positiveP ,
computed by Total Probability Formula.
test is positive 0.01 0.99 0.99 0.004 0.01386P
3) In the numerator of Bayes’ formula, we have to put the product that
corresponds with branch “Disease – Positive”. 0.01 0.99 0.0099
4) Finally,
0.0099
has disease given, test positive0.01
0 1386
.7 43P
STT 351 Conditional Probability. Independence 2.4 -2.5
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2.5. Independence.
Example2.5.2.
You toss a coin and it comes up “H” two times. What is the probability that the
next toss will also be a "H"?
Answer. 1
2
Example 2.5.3.
Are two mutually exclusive events independent or no? Answer. No
Example 2.5.4.
Consider throwing two fair dice, red and green. Are the following events
independent or dependent? Give reasons for your answers.
1) A = {the sum is 4} and B = {Both dice show the same number}
2) A = {the sum is 6} and B = {The red die shows an even}
Objectives.
• Definition of independence of two events
• Multiplication rule of probability of intersection of two events
• Independence of more than two events
2.5.1. Definition of Independence of Two Random Events.
Two events A and B are called independent if and only if
P A B P A P B
It can be easy shown that for independent events A and B
,P A B P A P A B P A
Independence of two events in the probability theory means that the chances of
occurrence of one even do not change when the info about occurrence of another
event is provided.
STT 351 Conditional Probability. Independence 2.4 -2.5
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Solution.
1) {sum is 4}={(1,3), (2,2), (3,1)}A 3
the sum is 436
1
12P
1 1
the sum is 4 dice show the same1
6 12 6P
Conditional probability ≠ Unconditional probability
A and B are dependent events
2)
{sum is 6}={(5,1), (4, 2), (3, 3), (2, 4), (1, 5)}
36P
5
A
A
3
36
1
12P A B
Conditional probability ≠ Unconditional probability
A and B are dependent events
STT 351 Conditional Probability. Independence 2.4 -2.5
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Example 2.5.5.
An ice cream company wishes to use red dye to enhance the color in its strawberry
ice cream. The Food and Drug Administration (FDA) requires the dye to be tested
for cancer-producing potential in a laboratory using laboratory rats. The results of
one test on 1,000 rats are summarized in the following table:
a) Convert the table into the proportions table by dividing each entry by 1,000.
b) What is the probability that a rat develops cancer, given that this rat ate red
dye? 60
500
30.12
25P C R
c) Are “developing cancer” and “eating dye” independent events?
0.06 0.02 0.08P C , P C R P C events are not independent
d) Should the FDA approve or ban the use of the dye?
Explain why or why not using 0.12 0.08P C R P C .
FDA should ban the use of Red Dye because eating this substance increases
Probability of developing cancer.
Developed
Cancer
C
No
Cancer
C
Totals
Ate Red Dye R 60 440 500
Did Not Eat Red Dye R 20 480 500
Totals 80 920 1,000
Developed
Cancer
C
No
Cancer
C
Totals
Ate Red Dye R 0.06 0.44 0.5
Did Not Eat Red Dye R 0.02 0.48 0.5
Totals 0.08 0.92 1
STT 351 Conditional Probability. Independence 2.4 -2.5
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2.5.6. An Important Application of Bayes’ Formula (for reading).
Computing the probability that a message containing a given word is spam.
Let's suppose the suspected message contains the word "replica". Most people who
are used to receiving e-mail know that this message is likely to be spam, a proposal
to buy counterfeit copies of well-known brands of watches. The spam detection
software, however, does not "know" such facts; all it can do is to compute
probabilities.
The formula used by the software to determine that, is derived from Bayes'
theorem
P S P W S
P S WP S P W S P H P W H
where:
• P S W is the probability that a message is a spam, knowing that the word
"replica" is in it;
• P S is the overall probability that any given message is spam;
• P W S is the probability that the word "replica" appears in spam messages;
• P H is the overall probability that any given message is not spam (is
"ham");
• P W H is the probability that the word "replica" appears in ham messages.
The spamicity of a word.
Recent statistic shows that the current probability of any message being spam is
80%. Thus,
0.8; 0.2P S P H
However, most Bayesian spam detection software makes the assumption that there
is no a priori reason for any incoming message to be spam rather than ham, and
considers both cases to have equal probabilities of 50%:
0.5; 0.5P S P H
STT 351 Conditional Probability. Independence 2.4 -2.5
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The filters that use this hypothesis are said to be "not biased", meaning that they
have no prejudice regarding the incoming email.
This assumption permits simplifying the general formula to:
P W S
P S WP W S P W H
This is functionally equivalent to asking, "what percentage of occurrences of the
word "replica" appear in spam messages?"
The quantity
P W S
P S WP W S P W H
is called "spamicity" (or "spaminess") of the word "replica", and can be computed.
The number P W S used in this formula is approximated to the frequency of
messages containing "replica" in the messages identified as spam during the
learning phase. Similarly, P W H is approximated to the frequency of messages
containing "replica" in the messages identified as ham during the learning phase.
For these approximations to make sense, the set of learned messages needs to be
big and representative enough. It is also advisable that the learned set of messages
conforms to the 50% hypothesis about repartition between spam and ham, i.e. that
the datasets of spam and ham are of same size.
Of course, determining whether a message is spam or ham based only on the
presence of the word "replica" is error-prone, which is why Bayesian spam
software tries to consider several words and combine their spamicities to determine
a message's overall probability of being spam.