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Probability: Definition
• Probability theory is the branch of mathematics that
studies the possible outcomes of given events
together with the outcomes’ relative likelihoods and
distributions.
• In common usage, the word “probability” is used to
mean the chance that a particular event (or set of
events) will occur expressed on a linear scale from 0
(impossibility) to 1 (certainty), also expressed as a
percentage between 0 and 100%.
• The analysis of data (possibly generated by
probability models) is called statistics.
• Find the probability of getting a 3 or 5 when throwing
a die.
Definitions
Definitions
Types of Probability
• Classical Probability:
• Empirical Probability:
Types of Probability
• The empirical probability definition has a
weakness that it depends on the results of a
particular experiment.
• The next time this experiment is repeated, you
are likely to get a somewhat different result.
• However, as an experiment is repeated many
times, the empirical probability of an event,
based on the combined results, approaches
the theoretical probability of the event.
Axioms of Probability
Example:
Probability that John passes a Math exam is 4/5 and that he
passes a Chemistry exam is 5/6. If the probability that he passes
both exams is 3/4, find the probability that he will pass at least
one exam.
What is the probability of getting a total of 7 or 11, when two
dice are rolled?
Multiplication Law
If one operation can be performed in n1 ways, and if
for each of these a second operation can be performed
in n2 ways, then the two operations can be performed
together in n1n2 ways.
How large is the sample space when a pair of dice is
thrown?
Permutation and Combination
Permutation: The term permutations refers to an arrangement
of objects when the order matters
From among ten employees, three are to be selected to travel to three out-
of-town plants A, B, and C, one to each plant. Since the plants are located
in different cities, the order in which the employees are assigned to the
plants is an important consideration. In how many ways can the
assignments be made?
Permutation and Combination
Combination: The term combination refers to the arrangement
of objects when order does not matter
In the previous example, suppose that three employees are to be selected
from among the ten available to go to the same plant. In how many ways
can this selection be made?
Examples
A package of six light bulbs contains 2 defective bulbs. If three bulbs are
selected for use, find the probability that none of the three is defective.
A university warehouse has received a shipment of 25 printers, of which 10
are laser printers and 15 are inkjet models. If 6 of these 25 are selected at
random to be checked by a particular technician, what is the probability that
exactly 3 of these selected are laser printers? At least 3 inkjet printers?
Partitions
Consider 10 engineers to be split into 3 groups to be assigned to 3 plants. If we
are to send 5 people to Plant A, 3 people to Plant B, and 2 people to Plant C,
then the total number of assignments is:
Conditional Probability and
Independence
• Conditional probability is a tool for dealing with additional
information.
• Conditional probability is the probability of an event occurring
given the knowledge that another event has occurred.
• The conditional probability of event A occurring, given that
event B has occurred is denoted by P(A|B) and is read
“probability of A given B”.
Conditional Probability and
Independence
• Independence: The events A and B are called (statistically)
independent if P(A ∩ B) = P(A)P(B)
• Another way to express independence is to say that the
knowledge of B occurring does not change our assessment of
P(A). This means that P(A|B) = P(A).
• The probability that a person is female given that he or she
was born in March is just the same as the probability that the
person is female.
• If events A1, A2, ..., Ak are independent, then P(A1A2...Ak) =
P(A1) × P(A2) × ... × P(Ak)
Example
Let A = {a family has two boys} and B = {a family of two has at least one boy}
Find P(A | B).
P(A) = 1/4
Possibilities = {GG,GB,BG,BB}
N(A int B) = 1
N(B) = 3
P(A|B) = 1/3
Three bits (0 or 1 digits) are transmitted over a noisy channel, so they will be
flipped independently with probability 0.1 each. What is the probability that
a) At least one bit is flipped b) Exactly one bit is flipped?
B1B2B3
a)
P(at least one bit is flipped) = 1 – P(No bits are flipped)
= 1 – (0.9)(0.9)(0.9) = 1 – (0.9)3 = 0.271
b) P(B1’B2B3) + P(B1B2’B3) + P(B1B2B3’) = (0.1)(0.9)(0.9) + (0.9)(0.1)(0.9) +
(0.9)(0.9)(0.1) = 0.243
Bayes Rule
Events B1, B2, . . . , Bk are said to be a partition of the sample
space S if the following two conditions are satisfied.
a) BiBj = for each pair i, j
b) B1 B2 · · · Bk = S
This situation often arises when the statistics are available in
subgroups of a population.
For example, an insurance company might know accident rates for
each age group Bi . This will give the company conditional
probabilities P(A | Bi) (if we denote A = event of accident).
Question: if we know all the conditional probabilities P(A | Bi),
how do we find the unconditional P(A)?
A = Accident
S = {Young  18 -35 = 40, Middle-aged  36-56
= 25, Aged57 onwards = 35}
P(A|Y) = 0.6 = P(A inter Y) / P(Y)
P(A|MA) = 0.2
P(A|Ages) = 0.4
P(Y) = 40/100 = 0.4; P(MA) = 0.25; P(Aged) =
0.35
P(A) = P(A inter Y) + P(A inter MA) + P(A inter
Ages) = 0.24 + 0.5 + 0.14 = 0.88
P(A inter Y) = P(A|Y).P(Y) = 0.6*0.4 = 0.24
P(A inter MA) = P(A|MA)P(MA) = 0.2*0.25 = 0.5
P(A inter Aged) = P(A|Aged) P(Aged) = 0.4*0.35
= 0.14
Bayes Rule
Consider a case when k = 2: The event A can be written as
the union of mutually exclusive events AB1 and AB2, that is
A = AB1 AB2 it follows that P(A) = P(AB1) + P(AB2)
If the conditional probabilities of P(A|B1) and P(A|B2) are
known, that is P(A|B1) = P(AB1)/ P(B1) and P(A|B2) =
P(AB2)/P(B2) ,then P(A) = P(A|B1)P(B1) + P(A|B2)P(B2).
Suppose we want to find probability of the form P(B1|A),
which can be written as P(B1|A) = P(AB1) P(A) = P(A|B1)
P(B1) P(A) , therefore P(B1|A) = P(B1)P(A|B1) P(B1)P(A|B1)
+ P(B2)P(A|B2)
Bayes Rule : Bayesian Network
P(B1|A) = P(B1 inter A) / P(A)
P(A) = P(A inter B1) + P (A inter B2) + P(A inter B3) = P(A|B1).P(B1) +
P(A|B2).P(B2) + P(A|B3).P(B3)
P(B1 inter A) = P(A|B1) . P(B1)
P(B1|A) = P(A|B1) . P(B1) / (P(A|B1).P(B1) + P(A|B2).P(B2) + P(A|B3).P(B3))
Example
At a certain assembly plant, three machines make 30%, 45%, and 25%,
respectively, of the products. It is known from the past experience that 2%,
3%, and 2% of the products made by each machine, respectively, are
defective. Now, suppose that a finished product is randomly selected.
a) What is the probability that it is defective?
b) If a product were chosen randomly and found to be defective, what is the
probability that it was made by machine 3?
D = Product is Defective;; P(M1) = 0.3, P(M2) = 0.45, P(M3) = 0.25; M1, M2,
M3 P(D|M1) = 0.02 , P(D|M2) = 0.03 , P(D|M3) = 0.02, P(D) = ? P(M3|D) = ?
P(D) = 0.0245
P(M3|D) = 0.204
1-Probability-Conditional-Bayes.pdf
1-Probability-Conditional-Bayes.pdf

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1-Probability-Conditional-Bayes.pdf

  • 1. Probability: Definition • Probability theory is the branch of mathematics that studies the possible outcomes of given events together with the outcomes’ relative likelihoods and distributions. • In common usage, the word “probability” is used to mean the chance that a particular event (or set of events) will occur expressed on a linear scale from 0 (impossibility) to 1 (certainty), also expressed as a percentage between 0 and 100%. • The analysis of data (possibly generated by probability models) is called statistics. • Find the probability of getting a 3 or 5 when throwing a die.
  • 4. Types of Probability • Classical Probability: • Empirical Probability:
  • 5. Types of Probability • The empirical probability definition has a weakness that it depends on the results of a particular experiment. • The next time this experiment is repeated, you are likely to get a somewhat different result. • However, as an experiment is repeated many times, the empirical probability of an event, based on the combined results, approaches the theoretical probability of the event.
  • 7. Example: Probability that John passes a Math exam is 4/5 and that he passes a Chemistry exam is 5/6. If the probability that he passes both exams is 3/4, find the probability that he will pass at least one exam. What is the probability of getting a total of 7 or 11, when two dice are rolled?
  • 8. Multiplication Law If one operation can be performed in n1 ways, and if for each of these a second operation can be performed in n2 ways, then the two operations can be performed together in n1n2 ways. How large is the sample space when a pair of dice is thrown?
  • 9. Permutation and Combination Permutation: The term permutations refers to an arrangement of objects when the order matters From among ten employees, three are to be selected to travel to three out- of-town plants A, B, and C, one to each plant. Since the plants are located in different cities, the order in which the employees are assigned to the plants is an important consideration. In how many ways can the assignments be made?
  • 10. Permutation and Combination Combination: The term combination refers to the arrangement of objects when order does not matter In the previous example, suppose that three employees are to be selected from among the ten available to go to the same plant. In how many ways can this selection be made?
  • 11. Examples A package of six light bulbs contains 2 defective bulbs. If three bulbs are selected for use, find the probability that none of the three is defective. A university warehouse has received a shipment of 25 printers, of which 10 are laser printers and 15 are inkjet models. If 6 of these 25 are selected at random to be checked by a particular technician, what is the probability that exactly 3 of these selected are laser printers? At least 3 inkjet printers?
  • 12. Partitions Consider 10 engineers to be split into 3 groups to be assigned to 3 plants. If we are to send 5 people to Plant A, 3 people to Plant B, and 2 people to Plant C, then the total number of assignments is:
  • 13. Conditional Probability and Independence • Conditional probability is a tool for dealing with additional information. • Conditional probability is the probability of an event occurring given the knowledge that another event has occurred. • The conditional probability of event A occurring, given that event B has occurred is denoted by P(A|B) and is read “probability of A given B”.
  • 14. Conditional Probability and Independence • Independence: The events A and B are called (statistically) independent if P(A ∩ B) = P(A)P(B) • Another way to express independence is to say that the knowledge of B occurring does not change our assessment of P(A). This means that P(A|B) = P(A). • The probability that a person is female given that he or she was born in March is just the same as the probability that the person is female. • If events A1, A2, ..., Ak are independent, then P(A1A2...Ak) = P(A1) × P(A2) × ... × P(Ak)
  • 15. Example Let A = {a family has two boys} and B = {a family of two has at least one boy} Find P(A | B). P(A) = 1/4 Possibilities = {GG,GB,BG,BB} N(A int B) = 1 N(B) = 3 P(A|B) = 1/3 Three bits (0 or 1 digits) are transmitted over a noisy channel, so they will be flipped independently with probability 0.1 each. What is the probability that a) At least one bit is flipped b) Exactly one bit is flipped? B1B2B3 a) P(at least one bit is flipped) = 1 – P(No bits are flipped) = 1 – (0.9)(0.9)(0.9) = 1 – (0.9)3 = 0.271 b) P(B1’B2B3) + P(B1B2’B3) + P(B1B2B3’) = (0.1)(0.9)(0.9) + (0.9)(0.1)(0.9) + (0.9)(0.9)(0.1) = 0.243
  • 16. Bayes Rule Events B1, B2, . . . , Bk are said to be a partition of the sample space S if the following two conditions are satisfied. a) BiBj = for each pair i, j b) B1 B2 · · · Bk = S This situation often arises when the statistics are available in subgroups of a population. For example, an insurance company might know accident rates for each age group Bi . This will give the company conditional probabilities P(A | Bi) (if we denote A = event of accident). Question: if we know all the conditional probabilities P(A | Bi), how do we find the unconditional P(A)?
  • 17. A = Accident S = {Young  18 -35 = 40, Middle-aged  36-56 = 25, Aged57 onwards = 35} P(A|Y) = 0.6 = P(A inter Y) / P(Y) P(A|MA) = 0.2 P(A|Ages) = 0.4 P(Y) = 40/100 = 0.4; P(MA) = 0.25; P(Aged) = 0.35 P(A) = P(A inter Y) + P(A inter MA) + P(A inter Ages) = 0.24 + 0.5 + 0.14 = 0.88 P(A inter Y) = P(A|Y).P(Y) = 0.6*0.4 = 0.24 P(A inter MA) = P(A|MA)P(MA) = 0.2*0.25 = 0.5 P(A inter Aged) = P(A|Aged) P(Aged) = 0.4*0.35 = 0.14
  • 18. Bayes Rule Consider a case when k = 2: The event A can be written as the union of mutually exclusive events AB1 and AB2, that is A = AB1 AB2 it follows that P(A) = P(AB1) + P(AB2) If the conditional probabilities of P(A|B1) and P(A|B2) are known, that is P(A|B1) = P(AB1)/ P(B1) and P(A|B2) = P(AB2)/P(B2) ,then P(A) = P(A|B1)P(B1) + P(A|B2)P(B2). Suppose we want to find probability of the form P(B1|A), which can be written as P(B1|A) = P(AB1) P(A) = P(A|B1) P(B1) P(A) , therefore P(B1|A) = P(B1)P(A|B1) P(B1)P(A|B1) + P(B2)P(A|B2)
  • 19. Bayes Rule : Bayesian Network P(B1|A) = P(B1 inter A) / P(A) P(A) = P(A inter B1) + P (A inter B2) + P(A inter B3) = P(A|B1).P(B1) + P(A|B2).P(B2) + P(A|B3).P(B3) P(B1 inter A) = P(A|B1) . P(B1) P(B1|A) = P(A|B1) . P(B1) / (P(A|B1).P(B1) + P(A|B2).P(B2) + P(A|B3).P(B3))
  • 20. Example At a certain assembly plant, three machines make 30%, 45%, and 25%, respectively, of the products. It is known from the past experience that 2%, 3%, and 2% of the products made by each machine, respectively, are defective. Now, suppose that a finished product is randomly selected. a) What is the probability that it is defective? b) If a product were chosen randomly and found to be defective, what is the probability that it was made by machine 3? D = Product is Defective;; P(M1) = 0.3, P(M2) = 0.45, P(M3) = 0.25; M1, M2, M3 P(D|M1) = 0.02 , P(D|M2) = 0.03 , P(D|M3) = 0.02, P(D) = ? P(M3|D) = ? P(D) = 0.0245 P(M3|D) = 0.204