ST401: Mathematical Statistics with Applications (7th Edition) –
Wackerly – Complete Probability Theory Exam Prep with Detailed
Rationales
Course Code: ST401
Course Name: Probability Theory
Topic: Foundational Probability, Combinatorics, Conditional Probability, Bayes'
Theorem, and Random Variables, Advanced Probability Theory, Combinatorics,
Continuous Distributions, Joint Multivariate Distributions, and Sampling Methods
Academic Year: 2026/2027
based on Mathematical Statistics with Applications by Wackerly, Mendenhall, and
Scheaffer. It provides a structured set of problems with detailed step-by-step
rationales covering foundational probability, combinatorics, conditional
probability, Bayes' theorem, and discrete and continuous random variables
Formula Reference Architecture
• Binomial Distribution
o Probability Mass / Density Function: P(X = x) = (n choose x) * p^x *
(1-p)^(n-x)
o Mean (E[X]): np
o Variance (V[X]): np(1-p)
• Poisson Distribution
, o Probability Mass / Density Function: P(X = x) = (lambda^x * e^-
lambda) / x!
o Mean (E[X]): lambda
o Variance (V[X]): lambda
• Uniform (Continuous) Distribution
o Probability Mass / Density Function: f(x) = 1 / (b-a), where a <= x <=
b
o Mean (E[X]): (a+b) / 2
o Variance (V[X]): (b-a)^
• Exponential Distribution
o Probability Mass / Density Function: f(x) = (1 / beta) * e^(-x/beta),
where x >= 0
o Mean (E[X]): beta
o Variance (V[X]): beta^2
Part I: Exam Prep Questions
Question 1
A committee of 3 members is to be selected from a pool of 5 mathematicians and 4
statisticians. What is the probability that the committee consists of exactly 2
mathematicians and 1 statistician?
A) 2/7
B) 5/14
C) 10/21
D) 5/9
• CORRECT ANSWER: C) 10/21
• RATIONALE: The total number of ways to choose any 3 members from
the 9 available is (9 choose 3) = (9 * 8 * 7) / (3 * 2 * 1) = 84. The number of
ways to choose exactly 2 mathematicians from 5 is (5 choose 2) = 10. The
number of ways to choose exactly 1 statistician from 4 is (4 choose 1) = 4.
, By the multiplication principle, the number of favorable outcomes is 10 * 4
= 40. Therefore, the probability is = .
Question 2
Consider a diagnostic test for a disease. 1% of the population actually has the
disease. The test has a false positive rate of 5% and a true positive rate (sensitivity)
of 99%. If a randomly selected individual tests positive, what is the probability that
they actually have the disease?
A) 1/6
B) 1/5
C) 99/100
D) 1/2
• CORRECT ANSWER: A) 1/6
• RATIONALE: Let D be the event of having the disease and T+ be the
event of testing positive. We are given P(D) = 0.01, so P(D^c) = 0.99. The
true positive rate is P(T+|D) = 0.99. The false positive rate is P(T+|D^c) =
0.05. By Bayes' Theorem: P(D|T+) = [P(T+|D)P(D)] / [P(T+|D)P(D) +
P(T+|D^c)P(D^c)] = (0.99 * 0.01) / [(0.99 * 0.01) + (0.05 * 0.99)] = 0.0099 /
(0.0099 + 0.0495) = 0..0594 = 1/6.
Question 3
A continuous random variable X has a probability density function given by f(x) =
c * x^2 for 0 <= x <= 2, and f(x) = 0 elsewhere. Find the value of the constant c
that normalizes this density function.
A) 1/4
B) 3/8
C) 1/2
D) 3/4
• CORRECT ANSWER: B) 3/8
• RATIONALE: For f(x) to be a valid probability density function, its
integral over the entire space must equal 1. The integral from 0 to 2 of c *
x^2 dx = c * [x^] evaluated from 0 to 2 = c * (8/3 - 0) = 8c/3. Setting
8c/3 = 1 yields c = 3/8.
Question 4
, An electronic component has a lifetime described by an exponential distribution
with a mean lifetime (beta) of 1000 hours. What is the probability that a
component lasts more than 2000 hours?
A) 1 - e^-2
B) e^-1
C) e^-2
D) 2e^-1
• CORRECT ANSWER: C) e^-2
• RATIONALE: The probability density function of an exponential random
variable is f(x) = (1/beta) * e^(-x/beta) for x >= 0. The survival function,
which gives the probability that X > x, is P(X > x) = integral from x to
infinity of (1/beta) * e^(-t/beta) dt = e^(-x/beta). Substituting x = 2000 and
beta = 1000, we obtain P(X > 2000) = e^(-2000/1000) = e^-2.
Question 5
Let X be a discrete random variable representing the number of traffic accidents at
an intersection per week, following a Poisson distribution with a mean (lambda) of
3. Find the probability that exactly 2 accidents occur in a given week.
A) 3e^-3
B) 4.5e^-3
C) 9e^-3
D) 2e^-3
• CORRECT ANSWER: B) 4.5e^-3
• RATIONALE: The probability mass function of a Poisson distribution is
P(X = x) = (lambda^x * e^-lambda) / x!. For lambda = 3 and x = 2, we find
P(X = 2) = (3^2 * e^-3) / 2! = (9 * e^-3) / 2 = 4.5e^-3.
Question 6
A total of 10 items are manufactured, out of which 3 are defective. If a sample of 2
items is chosen at random without replacement, find the probability that at least
one item is defective.
A) 7/15
B) 8/15
C) 1/3
D) 2/3
Wackerly – Complete Probability Theory Exam Prep with Detailed
Rationales
Course Code: ST401
Course Name: Probability Theory
Topic: Foundational Probability, Combinatorics, Conditional Probability, Bayes'
Theorem, and Random Variables, Advanced Probability Theory, Combinatorics,
Continuous Distributions, Joint Multivariate Distributions, and Sampling Methods
Academic Year: 2026/2027
based on Mathematical Statistics with Applications by Wackerly, Mendenhall, and
Scheaffer. It provides a structured set of problems with detailed step-by-step
rationales covering foundational probability, combinatorics, conditional
probability, Bayes' theorem, and discrete and continuous random variables
Formula Reference Architecture
• Binomial Distribution
o Probability Mass / Density Function: P(X = x) = (n choose x) * p^x *
(1-p)^(n-x)
o Mean (E[X]): np
o Variance (V[X]): np(1-p)
• Poisson Distribution
, o Probability Mass / Density Function: P(X = x) = (lambda^x * e^-
lambda) / x!
o Mean (E[X]): lambda
o Variance (V[X]): lambda
• Uniform (Continuous) Distribution
o Probability Mass / Density Function: f(x) = 1 / (b-a), where a <= x <=
b
o Mean (E[X]): (a+b) / 2
o Variance (V[X]): (b-a)^
• Exponential Distribution
o Probability Mass / Density Function: f(x) = (1 / beta) * e^(-x/beta),
where x >= 0
o Mean (E[X]): beta
o Variance (V[X]): beta^2
Part I: Exam Prep Questions
Question 1
A committee of 3 members is to be selected from a pool of 5 mathematicians and 4
statisticians. What is the probability that the committee consists of exactly 2
mathematicians and 1 statistician?
A) 2/7
B) 5/14
C) 10/21
D) 5/9
• CORRECT ANSWER: C) 10/21
• RATIONALE: The total number of ways to choose any 3 members from
the 9 available is (9 choose 3) = (9 * 8 * 7) / (3 * 2 * 1) = 84. The number of
ways to choose exactly 2 mathematicians from 5 is (5 choose 2) = 10. The
number of ways to choose exactly 1 statistician from 4 is (4 choose 1) = 4.
, By the multiplication principle, the number of favorable outcomes is 10 * 4
= 40. Therefore, the probability is = .
Question 2
Consider a diagnostic test for a disease. 1% of the population actually has the
disease. The test has a false positive rate of 5% and a true positive rate (sensitivity)
of 99%. If a randomly selected individual tests positive, what is the probability that
they actually have the disease?
A) 1/6
B) 1/5
C) 99/100
D) 1/2
• CORRECT ANSWER: A) 1/6
• RATIONALE: Let D be the event of having the disease and T+ be the
event of testing positive. We are given P(D) = 0.01, so P(D^c) = 0.99. The
true positive rate is P(T+|D) = 0.99. The false positive rate is P(T+|D^c) =
0.05. By Bayes' Theorem: P(D|T+) = [P(T+|D)P(D)] / [P(T+|D)P(D) +
P(T+|D^c)P(D^c)] = (0.99 * 0.01) / [(0.99 * 0.01) + (0.05 * 0.99)] = 0.0099 /
(0.0099 + 0.0495) = 0..0594 = 1/6.
Question 3
A continuous random variable X has a probability density function given by f(x) =
c * x^2 for 0 <= x <= 2, and f(x) = 0 elsewhere. Find the value of the constant c
that normalizes this density function.
A) 1/4
B) 3/8
C) 1/2
D) 3/4
• CORRECT ANSWER: B) 3/8
• RATIONALE: For f(x) to be a valid probability density function, its
integral over the entire space must equal 1. The integral from 0 to 2 of c *
x^2 dx = c * [x^] evaluated from 0 to 2 = c * (8/3 - 0) = 8c/3. Setting
8c/3 = 1 yields c = 3/8.
Question 4
, An electronic component has a lifetime described by an exponential distribution
with a mean lifetime (beta) of 1000 hours. What is the probability that a
component lasts more than 2000 hours?
A) 1 - e^-2
B) e^-1
C) e^-2
D) 2e^-1
• CORRECT ANSWER: C) e^-2
• RATIONALE: The probability density function of an exponential random
variable is f(x) = (1/beta) * e^(-x/beta) for x >= 0. The survival function,
which gives the probability that X > x, is P(X > x) = integral from x to
infinity of (1/beta) * e^(-t/beta) dt = e^(-x/beta). Substituting x = 2000 and
beta = 1000, we obtain P(X > 2000) = e^(-2000/1000) = e^-2.
Question 5
Let X be a discrete random variable representing the number of traffic accidents at
an intersection per week, following a Poisson distribution with a mean (lambda) of
3. Find the probability that exactly 2 accidents occur in a given week.
A) 3e^-3
B) 4.5e^-3
C) 9e^-3
D) 2e^-3
• CORRECT ANSWER: B) 4.5e^-3
• RATIONALE: The probability mass function of a Poisson distribution is
P(X = x) = (lambda^x * e^-lambda) / x!. For lambda = 3 and x = 2, we find
P(X = 2) = (3^2 * e^-3) / 2! = (9 * e^-3) / 2 = 4.5e^-3.
Question 6
A total of 10 items are manufactured, out of which 3 are defective. If a sample of 2
items is chosen at random without replacement, find the probability that at least
one item is defective.
A) 7/15
B) 8/15
C) 1/3
D) 2/3
