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Free Powerpoint Templates
EQT 272 PROBABILITY
AND STATISTICS
ROHANA BINTI ABDUL HAMIDINSTITUT E FOR ENGINEERING MATHEMATICS (IMK)
UNIVERSITI MALAYSIA PERLIS
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WHY DO COMPUTER ENGINEERS NEED TO STUDY PROBABILITY???????
1.Signal processing2.Computer memories3.Optical communication systems4.Wireless communication systems5.Computer network traffic
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Probability and statistics are related in an important way. Probability is used as a tool; it allows you to evaluate the reliability of your conclusions about the population when you have only sample information.
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Probability• Probability is a measure of the likelihood
of an event A occurring in one experiment
or trial and it is denoted by P (A).
)(
)(
)(outcomesofnumbertotal
)(occurcaneventthethatwaysofnumber)(
Sn
An
S
AAAP
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Experiment• An experiment is any process of
making an observation leading to outcomes for a sample space.
Example:
-Toss a die and observe the number that appears on the upper face.-A medical technician records a person’s blood type.-Recording a test grade.
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The mathematical basis of probability is the
theory of sets. • Sets
A set is a collection of elements or components• Sample Spaces, S
A sample space consists of points that correspond to all possible outcomes.
• Events
An event is a set of outcomes of an experiment and a subset of the sample space.
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• Experiment: Tossing a die• Sample space: S ={1, 2, 3, 4, 5, 6}• Events: A: Observe an odd number B: Observe a number less than 4 C: Observe a number which could divide by 3
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1. The union of events A and B, which is denoted as ,
- is the set of all elements that belong to A or B or both.
- Two or more events are called collective exhaustive events if the unions of these events result in the sample space.
2. The intersection of events A and B, which is denoted by ,
- is the set of all elements that belong to both A and B.
- When A and B have no outcomes in common, they are said to be mutually exclusive or disjoint sets.
3. The event that contains all of the elements that do not belong to an event A is called the complement of A and is denoted by
A B
A B
A
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Example 1.2
• Given the following sets;
A= {2, 4, 6, 8, 10}
B= {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}
C= {1, 3, 5, 11,….}, the set of odd numbers
Find , and BA BA C
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Answer
• = {1, 2, 3, 4, 5, 6, 7, 8, 9,10}
• = {2, 4, 6, 8, 10}
• = {2, 4, 6, 8,…}, the set of even numbers
BABA
C
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• A survey finds that 56% of people are married. They ask the same group of people, and 67% have at least one child. There are 41% that are married and have at least one child. Describe this results with a Venn diagram.
Example 1.3
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A group of 100 factory workers were questioned by a popular health magazine and 48% were found to take regular exercise. When asked about their eating habits, 67% replied that they always have breakfast. Not only that, 32% always have breakfast and exercise regularly. Describe this results with a Venn diagram.
Exercise 1
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1) 0 ( ) 1
2) ( ) ( ) 1
3) ( ) ( ) ( )
4) ( ) ( ) ( )
5) ( ) 1 ( )
6) (( ) ) ( )
7) (( ) ) ( )
8) ( ( )) ( )
9) ( ) [( ) ( )]
P A
P A P A
P A B P A P A B
P A B P B P A B
P A B P A B
P A B P A B
P A B P A B
P A A B P A B
P B P A B A B
A B A BA B
S
AB
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Theorem 1.1 : Laws of Probability
1 2
a) ( ) 1 –
b) ( ) – ( )
c) ( ) – ( ) – ( ) – ( ) ( )
d) If and are mutually exclusive events, then ( ) 0
e) If and are the subset of where
P A P A
P A B P A P B P A B
P A B C P A P B P C P A B P A C P B C P A B C
A B P A B
A A S A
1 2 , 1 2 then A P A P A
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Two fair dice are thrown. Determine
a) the sample space of the experiment
b) the elements of event A if the outcomes of both dice thrown are showing the same digit.
c) the elements of event B if the first thrown giving a greater digit than the second thrown.
d) probability of event A, P(A) and event B, P(B)
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Solutions 1.4
a) Sample space, S
1 2 3 4 5 6
1 (1, 1) (1, 2) (1, 3) (1, 2) (1, 5) (1, 6)
2 (2, 1) (2, 2) (2, 3) (2, 4) (2, 5) (2, 6)
3 (3, 1) (3, 2) (3, 3) (3, 4) (3, 5) (3, 6)
4 (4, 1) (4, 2) (4, 3) (4, 4) (4, 5) (4, 6)
5 (5, 1) (5, 2) (5, 3) (5, 4) (5, 5) (5, 6)
6 (6, 1) (6, 2) (6, 3) (6, 4) (6, 5) (6, 6)
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Solutions 1.4
b) A = {(1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6)}
c) B = {(2, 1), (3, 1), (3, 2), (4, 1), (4, 2), (4, 3), (5, 1), (5, 2), (5, 3),
(5, 4), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5)}
( ) 6 1d)
( ) 36 6
( ) 15 5
( ) 36 12
n AP A
n S
n BP B
n S
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Consider randomly selecting a UniMAP Master Degree
international student, and let A denote the event that the selected individual has a Visa Card and B has a Master Card. Suppose that P(A) = 0.5 and P(B) = 0.4 and = 0.25.
a) Compute the probability that the selected individual has at least one of the two types of cards ?
b) What is the probability that the selected individual has neither type of card?
( )P A B
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Solutions 1.5
a) ( ) – ( )
= 0.5 0.4 – 0.25 0.65
b) 1 ( ) 1– 0.65 0.35
P A B P A P B P A B
P A B
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• Some experiments can be generated in stages, and the sample space can be displayed in a tree diagram.
• Each successive level of branching on the tree corresponds to a step required to generate the final outcome.
• A tree diagram helps to find simple events.
1.4.1 Tree diagrams
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• A box contains one yellow and two red balls. Two balls are randomly selected and their colors recorded. Construct a tree diagram for this experiment and state the simple events.
Y1 R1
R2
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First ball Second ball RESULTS
Y1
R1
R2
R1
R2
Y1
R2
Y1
R1
Y1R1
Y1R2
R1Y1
R1R2
R2Y1
R2R1
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Exercise 2
• 3 people are randomly selected from voter registration and driving records to report for jury duty. The gender of each person is noted by the county clerk. List the simple events by creating a tree diagram.
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• We can use counting techniques or counting rules to
1.4.2 Counting technique
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• This counting rule count the number of outcomes when the experiment involves selecting r objects from a set of n objects when the order of selection is important.
)!(
!
rn
nPr
n
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• The number of ways to arrange
an entire set of n distinct items is
!nPnn
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• "The password of the safe was 472".
• We do care about the order. "724" would not work, nor would "247". It has to be exactly 4-7-2.
• To help you to remember, think "Permutation ... Position"
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• There are basically two types of permutation:
Repetition is Allowed: such as a lock. It could be "333".
No Repetition: for example the first three people in a running race. You can't be first and second.
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Repetition is Allowed
• When you have n things to choose from ... you have n choices each time!
• When choosing r of them, the permutations are:
• n × n × ... (r times)• (In other words, there are n possibilities for the
first choice, THEN there are n possibilites for the second choice, and so on, multplying each time.)
• Which is easier to write down using an exponent of r: n × n × ... (r times) = nr
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• Example:
• In a lock , there are 10 numbers to choose from (0,1,..9) and you choose 3 of them:
• 10 × 10 × ... (3 times)
= 103 = 1,000 permutations
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No Repetition
• In this case, you have to reduce the number of available choices each time.
• For example, what order could 16 pool balls be in?
• After choosing a ball, you can't choose it again.
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• So, your first choice would have 16 possibilites, and your next choice would then have 15 possibilities, then 14, 13, etc. And the total permutations would be:
• 16 × 15 × 14 × 13 × ... = 20,922,789,888,000
• But maybe you don't want to choose them all, just 3 of them, so that would be only:
• 16 × 15 × 14 = 3,360• In other words, there are 3,360 different
ways that 3 pool balls could be selected out of 16 balls.
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• Suppose you have 3 books, A, B and C but you have room for only two on your bookshelf. In how many ways can you select and arrange the two books when the order is important.
AA BB CC
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)!(
!
rn
nPr
n
6
)!23(
!32
3
P
There are 6 ways to select and arrange the books in order.
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Exercise 3
Three lottery tickets are drawn from a total of 50. If the tickets will be distributed to each of the employees in the order in which they are drawn, the order will be important. How many simple events are associated with the experiment?
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• This counting rule count the number of outcomes when the experiment involves selecting r objects from a set of n objects when the order of selection is not important.
!
! !
n nnCr r r n r
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• "My fruit salad is a combination of apples, grapes and bananas" We don't care what order the fruits are in, they could also be "bananas, grapes and apples" or "grapes, apples and bananas", its the same fruit salad.
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• Suppose you have 3 books, A, B and C but you have room for only two on your bookshelf. In how many ways can you select and arrange the two books when the order is not important.
AA BB CC
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3
)!23(!2
!32
3
C
There are 3 ways to select and arrange the books when the order is not important
!
! !
n nnCr r r n r
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Exercise 4
Suppose that in the taste test, each participant samples 8 products and is asked the 3 best products, but not in any particular order. Calculate the number of possible answer test.
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TRY!
• A company produces 10 microchips during a night staff. 6 of these turn out to be defective. Suppose 3 of the chips were sent to a customer.
a)What is the probability of the customer receiving 2 defective chips?
b)What is the probability no defective chips?
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• Definition:
For any two events A and B with P(B) > 0, the conditional probability of A given that B has occurred is defined by
( )( | )
( )
P A BP A B
P B
GIVEN
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A study of 90 students was done by UniMAP first year students. The results are given in the table :
Area/Gender Male (C) Female (D) Total
Urban (A) 35 10 45
Rural (B) 25 20 45
Total 60 30 90
If a student is selected at random and have been told that the individual is a male student, what is the probability of he is from urban area?
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Answer;
)|( CAP
)(
)(
CP
CAP
90/60
90/35
5833.0
Probability of male students from urban area
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In 2006, Edaran Automobil Negara (EON) will produce a multipurpose national car (MPV) equipped with either manual or automatic transmission and the car is available in one of four metallic colours. Relevant probabilities for various combinations of transmission type and colour are given in the accompanying table:
Transmission
type/ColourGrey (C) Blue
Black
(B)Red
Automatic, (A) 0.15 0.10 0.10 0.10
Manual 0.15 0.05 0.15 0.20
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• Let, A = automatic transmissionB = blackC = grey Calculate;
a) ( ), ( ) and ( )
b) ( | ) and ( | )
c) ( | ) and ( | )
P A P B P A B
P A B P B A
P A C P A C
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Answer;
a) P(A) = probability of MPV with automatic transmission
P(A) = 0.15+0.10+0.10+0.10 = 0.45
P(B) = probability of black MPV
P(B) = 0.10+0.15 = 0.25
P(A∩B) = probability of black MPV with automatic transmission
P(A∩B) = 0.10
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P(A|B) = probability of auto MPV given that the MPV is black
P(B|A) = probability of black MPV given that the MPV has automatic transmission
)(
)()|(
BP
BAPBAP
25.0
1.0 4.0
)(
)()|(
AP
BAPABP
45.0
1.0 222.0
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P(A|C) = probability of auto MPV given that the MPV is grey
P(A|C’) = probability of auto MPV given that the MPV is not grey
)(
)()|(
CP
CAPCAP
3.0
15.0 5.0
)(
)()|(
CP
CAPCAP
7.0
3.0 429.0
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1 2 If , ,..., is a partition of a sample space, then the
of events conditional on an event can be obtained
from the probabilities and | using the formula,
|
n
i
i i
i
A A A
A B
P A P B A
P A
posterior
probabilities
1
| |
|
i i i i in
j jj
P A B P A P B A P A P B AB
P B P B P A P B A
-Used to revise previously calculated probabilities based on new information.
-Extension of conditional probability
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• Suppose someone told you they had a nice conversation with someone on the train. Not knowing anything else about this conversation, the probability that they were speaking to a woman is 50%.
• Now suppose they also told you that this person had long hair. It is now more likely they were speaking to a woman, since women are more likely to have long hair than men.
• Bayes' theorem can be used to calculate the probability that the person is a woman.
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• Suppose it is also known that 75% of women have long hair. Likewise, suppose it is known that 15% of men have long hair.
• Our goal is to calculate the probability that the conversation was held with a woman, given the fact that the person had long hair.
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• Probability that the conversation was held with a woman, given the fact that the person had long hair
)|( longwomenP
)()|()()|(
)()|(
manPmanlongPwomenPwomenlongP
womenPwomenlongP
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5.0)(,5.0)( manPwomenP
15.0)|(
,75.0)|(
manlongP
womenlongP
)()|()()|(
)()|(
manPmanlongPwomenPwomenlongP
womenPwomenlongP
)5.0(15.0)5.0(75.0
)5.0(75.0
83.0
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A drilling company has estimated a 40% chance of striking oil for their new well. A detailed test has been scheduled for more information. Historically, 60% of successful wells have had detailed tests, and 20% of unsuccessful wells have detailed tests. Given that this well has been scheduled for a detailed test, what is the probability that the well will be successful?
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• Answer:
)detailed|successful(P
)failure()failure|detailed()success()success|detailed(
)success()success|detailed(
PPPP
PP
6.0)(,4.0)( failurePsuccessP
2.0)failure|detail(
,6.0)success|detail(
P
P
)6.0(2.0)4.0(6.0
)4.0(6.0
6667.0
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TRY!!!• You have a database of 100 emails. • 60 of those 100 emails are spam
– 48 of those 60 emails that are spam have the word "buy"
– 12 of those 60 emails that are spam don't have the word "buy"
• 40 of those 100 emails aren't spam – 4 of those 40 emails that aren't spam have
the word "buy"– 36 of those 40 emails that aren't spam don't
have the word "buy"• What is the probability that an email is spam if it
has the word "buy"?
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• Definition : Two events A and B are said to be independent
if and only if either
Otherwise, the events are said to be dependent.
( | ) ( )
or
( | ) ( )
P A B P A
P B A P B
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Multiplicative Rule of Probability:
The probability that both two events and , occur is
( ) |
|
If and are independent,
( )
A B
P A B P A P B A
P B P A B
A B
P A B P A P B
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• Two events, A and B, are independent if the fact that
A occurs does not affect the
probability of B occurring.
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• Some other examples of independent events are:
Landing on heads after tossing a coin AND rolling a 5 on a single 6-sided die.
Choosing a marble from a jar AND landing on heads after tossing a coin.
Choosing a 3 from a deck of cards, replacing it, AND then choosing an ace as the second card.
Rolling a 4 on a single 6-sided die, AND then rolling a 1 on a second roll of the die.