AMS 110 Probability HW

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Probability Please read sections 3.1 – 3.3 in your textbook

Def: An experiment is a process by which observations are generated.

Def: A variable is a quantity that is observed in the experiment.

Def: The sample space (S) for an experiment is the set of all possible outcomes.

Def: An event E is a subset of a sample space. It provides the collection of outcomes that correspond to some classification.

Example:

 

 

 

 

 

 

 

Note: A sample space does not have to be finite.

Example: Pick any positive integer. The sample space is countably infinite.

A discrete sample space is one with a finite number of elements, { }1,2,3,4,5,6 or one that has a countably infinite number of elements { }1,3,5,7,... .

A continuous sample space consists of elements forming a continuum. { }x / 2 x 5< <

 

 

 

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A Venn diagram is used to show relationships between events.

 

A intersection B = (A ∩ B) = A and B

The outcomes in (A intersection B) belong to set A as well as to set B.

 

 

 

 

 

A union B = (A U B) = A alone or B alone or both

 

Union Formula

For any events A, B, P (A or B) = P (A) + P (B) – P (A intersection B) i.e. P (A U B) = P (A) + P (B) – P (A ∩ B)

 

 

 

 

 

 

 

 

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cA complement not A A ' A A = = = = A complement consists of all outcomes outside of A.

Note: P (not A) = 1 – P (A)

Def: Two events are mutually exclusive (disjoint, incompatible) if they do not intersect, i.e. if they do not occur at the same time. They have no outcomes in common.

 

 

 

 

When A and B are mutually exclusive, (A ∩ B) = null set = Ø, and P (A and B) = 0.

Thus, when A and B are mutually exclusive, P (A or B) = P (A) + P (B) (This is exactly the same statement as rule 3 below)

 

Axioms of Probability

Def: A probability function p is a rule for calculating the probability of an event. The function p satisfies 3 conditions:

1) 0 ≤ P (A) ≤1, for all events A in the sample space S

2) P (Sample Space S) = 1

3) If A, B, C are mutually exclusive events in the sample space S, then P(A B C) P(A) P(B) P(C)∪ ∪ = + +

 

 

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The Classical Probability Concept: If there are n equally likely possibilities, of which one

must occur and s are regarded as successes, then the probability of success is s n

.

Example:

 

Frequency interpretation of Probability: The probability of an event E is the proportion of times the event occurs during a long run of repeated experiments.

Example:

 

Def: A set function assigns a non-negative value to a set.

Ex: N (A) is a set function whose value is the number of elements in A.

Def: An additive set function f is a function for which f (A U B) = f (A) + f (B) when A and B are mutually exclusive.

N (A) is an additive set function.

Ex: Toss 2 fair dice. Let A be the event that the sum on the two dice is 5. Let B be the event that the sum on the 2 dice is 6.

N(A) = 4 since A consists of (1,4), (2,3), (3,2), (4,1).

N (B) = 5 since B consists of (1,5), (2,4), (3,3), (4,2), (5,1)

N (A or B) = 4 + 5 = 9

 

 

 

 

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DeMorgan’s Laws:

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

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Example: Problem 3.33, Pg. 74 Miller and Freund

In a group of 160 graduate engineering students, 92 are enrolled in statistics, 63 are enrolled in Operations Research and 40 are enrolled in both.

a. How many are enrolled in at least one course?

 

 

 

b. How many are enrolled in neither?

 

 

 

 

Example: Doctor Jackson makes the following observation of new clients entering a weight loss seminar:

80% are considered overweight [O], (need to lose at least 20 pounds), and will be placed on a low carbohydrate diet 70% do not exercise regularly, [NE] and will be joining a weight training class. 55% are both overweight and do not exercise regularly.

a) Draw a Venn diagram.

 

 

b) Are the events O and NE mutually exclusive?

c) Find the probability that a client selected at random either needs to lose weight or does not exercise regularly.

 

 

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d) Find the probability that a client selected at random is not overweight. We use the fact P (not A) = 1 – P (A)

 

 

e) Find the probability that a client selected at random is overweight but (and) exercises regularly

 

 

 

 

f) Find the probability that a client is not overweight and exercises regularly.

 

 

 

 

 

 

Def: Two events are independent if the occurrence or non-occurrence of one does not change the probability that the other will occur.

Example of Independent events: You roll 2 dice. Find the probability of rolling a 5 on each die.

 

 

If events A and B are independent, then P (A ∩ B) = P (A). P (B)

 

 

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Example of dependent events: Draw 2 cards from a standard deck of 52 cards without replacement. Find the probability of drawing 2 Aces.

If events A and B are dependent, then P (A ∩ B) ≠ P (A). P (B)

Laws of Probability: Multiplication Law of Probability: If A, B, C, and D are mutually independent events, then P (A ∩ B ∩ C ∩ D) = P (A).P (B).P(C).P (D) Addition Law of Probability:

For any events A and B: P (A U B) = P (A) + P (B) – P (A ∩ B)

For A, B, mutually exclusive: P (A U B) = P (A) + P (B)

For A, B, independent: P (A U B) = P (A) + P (B) – P (A). P (B)

P (A U B) = P (A) + P (B) [1 – P (A)]

 

 

 

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Conditional Probability: We use the notation P (A/B) to refer to the conditional probability that event A occurs, given the fact that event B is known to be true. Ex: The probability that someone who is known to be overweight is hypertensive would be written as P (Hypertensive/Overweight) = P (H/O).

P(A B)P(A / B) P(B) ∩

=

Venn diagram: Given the fact that B has already occurred, makes the new sample space the outcomes in B. Thus, P (A/B) = the ratio of the outcomes in the intersection of A and B to the outcomes in B.

Similarly, P (B/A)= P(B A) P(A) ∩ = P(A B)

P(A) ∩

Venn diagram: Conditional probabilities can also be thought of as limiting the type of people you want to include in your probability assessment.

 

 

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Two events are independent if knowing that one occurs does not change the probability that the other occurs. 3 ways to check if two events A and B are independent: 1. If P (A/B) = P (A), then they are independent. 2. If P (B/A) = P (B), then they are independent. 3. If P (A ∩ B) = P (A). P (B) If any of these is true, the others are also true. If A, B are independent, cP(A /B) P(A) P(A /B )= = P (B/A) = P (B) = cP(B/ A ) If A, B are dependent, P (A/B) ≠ P (A) ≠ cP(A /B ) P (B/A) ≠ P (B) ≠ cP(B/ A ) P (A ∩ B) ≠ P(A) P(B) If we know the probabilities of any two events as well as the probability of their intersection we can calculate both P (A/B) as well as P (B/A). In addition, for any 2 events A, B, if we know the probabilities of two events and the corresponding conditional probabilities we can calculate the probability of their intersection P (A and B) as follows: P (A ∩ B) = P (B/A) P(A) = P(A/B) P(B) This is known as the multiplication rule.

 

 

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Example: Current estimates are that about 25% of all deaths are due to cancer, and of the deaths due to cancer, 30% are attributed to tobacco, 40% to poor diet and 30% to other causes. Find the probability that a death is due to cancer and tobacco. Please use the multiplication rule.

 

When A and B are mutually exclusive, are they dependent or independent?

Answer: They are _______________, since

 

 

 

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Example:

P (Smoker) = .44 P (Chronic Cough) = .4 P (Smoker and Chronic Cough) = .37 a) Find the probability that a person is a smoker given the fact that he has chronic cough.

 

 

 

 

b) Find the probability that a person has a chronic cough given the fact that he is a smoker. P (CC/S) = c) Are the events S and CC dependent or independent?

 

 

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Example: In an experiment to study the dependence of hypertension on smoking habits, the following data were (datum: singular, data: plural) collected on 180 individuals: Smoking

Status

Nonsmoker Moderate Smoker

Heavy Smoker

Total

Hypertension Status

Hypertension 21 36 30

No Hypertension

48 26 19

Total

 

 

a. What is the probability that a randomly selected individual is experiencing hypertension?

 

b. Given that a heavy smoker is selected at random from this group, what is the probability that the person is experiencing hypertension?

 

 

c. Are the events hypertension and heavy smoker independent? Give 3 different supporting calculations.

 

 

 

 

 

 

d. Find the probability that a randomly selected individual is either a moderate smoker or is hypertensive.

 

 

 

 

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Smoking Status

 

Nonsmoker Moderate Smoker

Heavy Smoker

Total

Hypertension Status

Hypertension 21 36 30

No Hypertension

48 26 19

Total

 

 

e. Find the probability that a randomly selected individual is either a nonsmoker or a heavy smoker (OR, thus we use the union formula)

 

 

f. Find the probability that a randomly selected individual is not hypertensive but (but means AND) is a heavy smoker.

 

 

 

 

 

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Example:

Consider two events, A and B, of a sample space such that P (A) = P (B) = .6

a. Is it possible that the events A and B are mutually exclusive? Explain.

 

 

 

 

b. If the events A and B are independent, find the probability that the two events occur together, i.e., P (A and B)

 

 

 

c. If A and B are independent, find the probability that at least one of the two events will occur, i.e., P (A or B)

 

 

 

d. Suppose P (B/A) = .5, in this case are A and B independent or dependent?

 

 

 

 

e. Suppose P (B/A) = .5. Find the P (A or B)

 

 

 

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Example:

For two events A and B, P (A) = .6, P (B) = .4, P (B/A) = .6

a. Are A and B independent or dependent?

 

 

 

b. Find P (A/B)

 

 

 

 

c. At least one, i.e. P (A or B)

 

 

 

d. P(not A and not B)

 

 

 

 

e. P (A but not B)

 

 

 

 

 

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Def: In probability theory, a set of events is collectively exhaustive if it incldes all possible outcomes.

Ex: A fair die is rolled. The outcomes 1, 2, 3, 4, 5, and 6 are collectively exhaustive, because they encompass the entire range of possible outcomes.

Bayes Theorem

Def: Events 1 2 kB , B , .B … are said to partition a sample space S if two conditions exist:

a. i jB B ∩ = ∅ = the null set for any pair i and j

 

b. 1 2 kB U B U .U B … = Sample Space S

 

 

 

 

 

 

 

 

Bayes Theorem: If 1 2 kB , B , .B … partition S, and if A is any event in the sample space S, then

k

i 1

P(A / Bj).P(Bj)P(Bj / A) P(A / Bi).P(Bi)

=

=

∑

 

 

 

 

B1 B2 B3 B4

 

 

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Further explanation:

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

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Example of Bayes Theorem: In New York City, 51% of all adults are males. One masked adult was seen holding up an Apple Bank branch.

a) What is the probability that the person is a male?

 

 

 

 

b) It is known that immediately after leaving the bank branch, the person was smoking a cigar. Also, it is known that 9.5% of males smoke cigars whereas 1.7% of females smoke cigars based on data from the Substance Abuse and Mental Health Services Administration. Use this additional information to find the probability that the person holding up the bank was a female.

 

 

 

 

 

 

c) If the person was smoking a cigar, what is the probability that the person who held up the bank is male?

 

 

 

d) Use a table to obtain the same answers.

 

 

 

 

 

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A screening or diagnostic test for some condition is Positive (+T) if it states that the condition is present and negative (-T) if it indicates that the condition is absent.

The validity of a test is defined as the ability of a test to distinguish between those who have the disease and those who do not. Validity has 2 components:

1. The sensitivity of the test is defined as the ability to correctly identify those who have the disease. i.e. P(+T/D)

2. The specificity of the test is defined as the ability to correctly identify those who do not have the disease. P(-T/D’)

Note: In order to calculate the sensitivity and specificity of a test, we must know who really has the disease and who does not from some other source than the diagnostic test. This external source of truth is the gold standard, identifying the disease status of each member of the population.

The prevalence of a disease is the probability of currently having the disease. It is the (# of people who currently have the disease) ÷ (# of people in the population)

Term Brief Definition Notation Sensitivity The probability of a positive test result given

that the individual tested actually has the disease

P( +T | +D )

Specificity The probability of a negative test result given that the individual tested does not have the disease

P( −T | −D )

P(False Positive) Probability of a positive test result given the individual does not have the disease

P( +T | −D )

P(False Negative) Probability of a negative test result given the individual does have the disease

)|( +− DTP

Prevalence Proportion of individuals that have a disease at a given point in time

)( +DP

Positive predictive value

Probability of disease given a positive test result )|( ++ TDP

Negative predictive value

Probability of not having disease given a negative test result

)|( −− TDP

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Example: Air traffic controllers are required to undergo random drug testing. A urine test detects the presence of amphetamines and barbiturates. The sensitivity of the test is known to be .96 and the specificity is .93. Based on past drug testing of air traffic controllers, the FAA reports that the probability of drug use at a given time is .007.

Tree Diagram for Air traffic controllers

Contingency Table for Air traffic controllers

a. What is the probability that a randomly selected air traffic controller will test positive? (If the test comes back positive, the individual will have to undergo a second test that is more accurate, but more expensive than the urine test.)

b. Find the probability that a person truly uses drugs, given that his test is positive, i.e., find the positive predictive value.

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c. Find the probability that the individual is not a drug user given that the test result is negative i.e. negative predictive value

d. How does the positive predictive value change if the prevalence rate increases to 15%?

e. How does the negative predictive value change if the prevalence is 15%?

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Example: Typical values reported for the mammogram which is used to detect breast cancer are sensitivity = .86, specificity = .88. Of the women who undergo mammograms at any given time, about 1% is estimated to actually have breast cancer.

Tree Diagram for Mammogram

Contingency Table for Mammogram

A. Prevalence = .01

a. Find the probability of a positive test

b. Of the women who receive a positive mammogram, what proportion actually have breast cancer?

c. If a woman tests negative, what is the probability that she does not have breast cancer?

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B. The prevalence rate for all women undergoing a mammogram was changed to .001

Tree diagram for young women

Contingency table for young women

a. Find the probability that a randomly selected young woman who tests positive actually has breast cancer.

b. Find the probability that a randomly selected young woman who tests negative does not have the disease.