Module 1: Probability
Basic Concepts of Probability
In considering probability, we deal with procedures (such as taking a polygraph test, rolling a die, answering a multiple-choice test question, or undergoing a test for drug use) that produce outcomes.
An event is any collection of results or outcomes of a procedure.
A simple event is an outcome or an event that cannot be further broken down into simpler components.
The sample space for a procedure consists of all possible simple events.
A compound event is any event combining two or more simple events.
Example: Birth of baby (f: female baby and m: male baby)
Procedure
Example of Event
Complete Sample Space
Single birth
1 female (simple event)
{f, m}
2 births
ff, mm, fm, mf
{ff, mm, fm, mf}
Notation for Probabilities:
Notation for Probabilities P denotes a probability.
A, and B denote specific events.
P(A) denotes the probability of event A occurring.
· Relative Frequency Approach:
P(A)= number of times A occurred/number of times the trial was repeated
· The Classical Approach:
P(A) = number of ways A can occur/number of distinct elements in the sample space
· Subjective Probability:
knowledge and experience from similar circumstances are used to predict the probability
· Properties of Probability:
1.
1.
1. P (an impossible event) = 0
2. P (a sure event) = 1
3. 0 ≤ P(E) ≤ 1
Example: Genotypes when studying the affect of heredity on height, we can express each individual genotype, AA, Aa, aA, and aa, on an index card and shuffle the four cards and randomly select one of them. What is the probability that we select a genotype in which the two components are different?
The sample space (AA, Aa, aA, aa) in this case includes equally likely outcomes. Among the 4 outcomes, there are exactly 2 in which the two components are different: Aa and aA. We can use the classical approach to get
P (outcome with different components) = 24 = 0.5
Summary of Key Probability Rules
Case
Probability
Observations
Special Addition Rule
P(A or B) = P(A) + P(B)
A and B are mutually exclusive
General Addition Rule
P(A or B) = P(A) + P(B) – P(A and B)
Always true
Special Multiplication Rule
P(A and B) = P(A) x P(B)
Only if A and B are independent
General Multiplication Rule
P(A and B) = P(A) x P(B/A)
Always true
Conditional Probability
P(B/A) = P(A and B) / P(A)
· A and B are mutually exclusive if they cannot occur simultaneously.
· A and B are independent events if the occurrence of one does not change the probability that the other occurs.
Examples:
The following table shows the results from experiments with polygraph instruments:
Test Result
The subject did not lie (No)
The subject lied (Yes)
Positive test result(subject lied)
15 (false positive)
42 (true positive)
Negative test result(subject did not lie)
32 (true negative)
9 (false negative)
a. If 1 subject is randomly probability of selecting a subject who had a positive test result or lied.
From the table we see that there are 66 subjects who had a positive test result or lied. We obtain that total of 66 by adding the subjects who tested positive to the subjects who lied, being careful to count everyone only once (15 + 42 + 9 = 66).
Dividing the total of 66 by the overall total of 98, we get: 66/98 or 0.673.
b. Assuming that 1 subject is randomly selected from the 98 that were tested, find the probability of selecting a subject who had a negative test result or did not lie.
From the table we must find the total number of subjects who had negative test results or did not lie, but we must find that total without double-counting. We get a total of 56 (from 32 + 9 + 15). Because 56 subjects had negative test results or did not lie, and because there are 98 total subjects included, we see P (negative test result or did not lie) = 5698 = 0.571
c. If two of the subjects included in the table are randomly selected without replacement, find the probability that the first selected person had a positive test result and the second selected person had a negative test result.
First selection:
P (positive test result) = 5798
(because there are 57 subjects who tested positive, and the total number of subjects is 98).
Second selection:
P (negative test result) = 4197
(after the first selection of a subject with a positive test result, there are 97 subjects remaining, 41 of whom had negative test results).
With P(first subject has positive test result) = 57/98 and P(second subject has negative test result) = 41/97, we have
P(1st subject has positive test result and 2nd subject has negative result) = 5798⋅4197=0.246
d. If 1 of the 98 test subjects is randomly selected, find the probability that the subject actually lied, given that he or she had a positive test result. That is, find P(subject lied | positive test result).
This is the probability that the selected subject lied, given that the subject had a positive test result. If we assume that the subject had a positive test result, we are dealing with the 57 subjects in the first row of Table 4-1. Among those 57 subjects, 42 lied, so
P(subject lied / positive test result) = 4257 = 0.737
Bayes Theorem
It is used to find a probability when other probabilities are known:
P(A|B) = P(A) P(B|A)P(B)
P(A|B): is the probability that A happens given that B happens
P(B|A): is the probability that B happens given that A happens
P(A): is the probability that A happens
P(B): is the probability that B happens
Example:
Two workers assemble parts from a production process. The probability that worker A makes a mistake in assembling a part is .02 and the probability that worker B makes a mistake is .03. However, worker A assembles 55% of the parts while worker B assembles the remaining 45%. If an assembled part is randomly selected from all those produced during a given time period and it is determined to be defective, what is the probability that worker A assembled this part?
A: worker A
B: worker B
D: defective
P(A/D) = P(A) P(D/A) / [P(A)P(D/A) + P(B)P(D/B)=
.55 x .02 / [.55 x .02 + .45 x .03] = .011/.025 = .449
References
Textbooks (Suggested)
Rajaretnam, T. (2016). Statistics for social sciences. Sage Publications, Inc. ISBN-13: 9789351506560
Triola, M. F. (2018). Elementary statistics (13th ed.). Pearson. ISBN-13: 978-0134462455 https://librarylogin-carolina.uagm.edu/login?url=https://search.ebscohost.com/login.aspx?direct=true&db=e000xww&AN=1214457&site=ehost-live&ebv=EB&ppid=pp_CoverLinks to an external site.
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