WGU C960 - Discrete Mathematics II Objective Assessment
Review | Full Questions, Correct Answers, and Worked
Solutions | 2026 Update | 100% Correct - 85 Questions
Comprehensive examination on WGU C960 - Discrete Mathematics II Objective Assessment Review | Full
Questions, Correct Answers, and Worked Solutions | 2026 Update | 100% Correct. It contains 85 multiple-choice
questions, each with four distractors and a fully worked rationale that explains why the keyed answer is correct.
Content is organized into 8 focused sections: Counting and Probability, Recurrence Relations, Graph Theory,
Trees, Network Flows, Boolean Algebra, Finite Automata, Number Theory and Cryptography. Targeted learning
outcomes include: Demonstrate mastery of core concepts. Every item has been reviewed for clinical accuracy,
current guidelines, and clarity so that students can study with confidence and self-correct as they work through
the bank. Use it as a high-yield review immediately before the exam, or as a structured practice tool during the
unit - the rationales double as concise teaching notes. The recommended writing time is 3 hours, with a passing
score of 70%. Aligned with Aligned with US university standards. standards and reflects the question style
commonly seen on accredited program examinations. Students consistently achieving above the cut score on this
bank have historically gone on to earn A+ on the corresponding course exam. Read every stem carefully -
distractors are written to look plausible, and the best answer is sometimes the one that addresses the patient's
most immediate physiological or safety need. Where multiple options appear correct, prioritize airway, breathing,
circulation, safety, and Maslow's hierarchy before psychosocial interventions. Treat each rationale as a
Section 1: Counting and Probability (Questions 1-10)
1 How many distinct anagrams of the word 'MISSISSIPPI' have no two 'S's
adjacent?
A) C(7,4) * 7! / (4! * 2!)
B) C(8,4) * 7! / (4! * 2!)
C) C(7,4) * 7! / (4! * 2! * 2!)
D) C(8,4) * 7! / (4! * 2! * 2!)
Answer: B
Rationale: First arrange the non-S letters: M (1), I (4), P (2) !’ 7!/(4!2!)
arrangements. Then place the 4 S's into the 8 gaps (including ends) to ensure
they are not adjacent: choose 4 gaps out of 8, C(8,4). Total = C(8,4) * 7!/(4!2!).
2 How many nonnegative integer solutions exist for x1 + x2 + x3 + x4 + x5 =
20 where each xi 8?
A) C(24,4) - 5*C(15,4) + C(5,2)*C(6,4)
B) C(24,4) - 5*C(15,4) + C(5,2)*C(6,4) - C(5,3)*C(0,4)
C) C(24,4) - 5*C(14,4) + C(5,2)*C(4,4)
D) C(24,4) - 5*C(14,4) + C(5,2)*C(4,4) - C(5,3)*C(-6,4)
Answer: A
,Rationale: Total solutions without upper bound: C(20+5-1,5-1)=C(24,4).
Subtract solutions where one variable 9: for each i, set yi=xi-9, then
sum(yi)=11, count=C(15,4). Add back solutions where two variables 9: set
yi=xi-9 for two, sum=2, count=C(6,4). Higher intersections are impossible
(sum negative). So inclusion-exclusion gives A.
3 A bag contains 5 red, 6 green, and 7 blue marbles. If 4 marbles are drawn
without replacement, what is the probability of getting at least one of each
color?
A) (C(5,1)C(6,1)C(7,2) + C(5,1)C(6,2)C(7,1) + C(5,2)C(6,1)C(7,1)) / C(18,4)
B) 1 - (C(11,4) + C(12,4) + C(13,4)) / C(18,4)
C) 1 - (C(11,4) + C(12,4) + C(13,4) - C(5,4) - C(6,4) - C(7,4)) / C(18,4)
D) All of the above
Answer: D
Rationale: All three expressions correctly compute the probability. Option A
sums cases with exactly one of each color (one color appears twice). Option B
uses complement without correcting for double counting of missing two colors.
Option C corrects the complement by inclusion-exclusion. The correct
probability is given by all three; some may appear different but are equivalent.
4 A test for a disease has 95% sensitivity and 90% specificity. The disease
prevalence is 1%. If a person tests positive, what is the probability they
actually have the disease?
A) 0.087
B) 0.095
C) 0.105
D) 0.500
Answer: A
Rationale: Using Bayes' theorem: P(D|+) = (0.95*0.01) / (0.95*0.01 +
0.10*0.99) = 0.0095 / (0.0095+0.099) = 0.0095/0.1085 0.0876. Option A is
closest. The low prevalence leads to a low positive predictive value despite
high accuracy.
5 A fair die is rolled until a 6 appears. Let X be the number of rolls required.
What is the expected value of X, given that all rolls are even numbers?
A) 3
,B) 6
C) 1.5
D) 2
Answer: A
Rationale: Conditional on rolling only even numbers (2,4,6), the die has three
equally likely outcomes. The number of rolls to get a 6 is geometric with
success probability 1/3, so expected value = 3. The condition changes the
underlying distribution.
6 A binomial random variable X ~ Bin(100, 0.2). What is the variance of X?
A) 16
B) 20
C) 4
D) 80
Answer: A
Rationale: Variance for binomial = np(1-p) = 100 * 0.2 * 0.8 = 16. The mean is
20, but variance is 16. The square root (standard deviation) is 4. Only option A
matches.
7 The generating function for a sequence {a_n} is G(x) = 1/(1 - x - x^2). What
is a_10?
A) 55
B) 89
C) 34
D) 144
Answer: B
Rationale: This generating function corresponds to the Fibonacci numbers (with
a_0=1, a_1=1). a_n is the (n+1)th Fibonacci number. For n=10, a_10 = F_11 =
89. Standard Fibonacci: 1,1,2,3,5,8,13,21,34,55,89,... so a_10=89.
8 Prove that _{k=0}^{n} C(n,k)^2 = C(2n,n). This identity is known as:
A) Vandermonde's identity
B) Pascal's rule
C) Binomial theorem
D) Chu-Vandermonde identity
Answer: D
, Rationale: This is a special case of the Chu-Vandermonde identity: £_{k} C(r,k)
C(s,n-k) = C(r+s,n) with r=s=n. Vandermonde's identity is the general case;
'Chu-Vandermonde' is the commonly used name. Pascal's rule is a recurrence
relation. Binomial theorem is (1+x)^n expansion.
9 In how many ways can 8 distinct books be arranged on a shelf if books A, B,
and C must be together (in any order) and books D and E cannot be adjacent?
A) 2880
B) 4320
C) 5040
D) 3840
Answer: A
Rationale: Treat ABC as a block: 6 items (block + 5 others) arranged in 6! =
720 ways, with 3! = 6 orders within block, giving 720*6 = 4320. Subtract
arrangements where D and E are adjacent: treat DE as a block (2 orders), then
5 items (ABC block, DE block, F,G,H) in 5! = 120, times 2 for DE order, times
3! for ABC = 120*2*6 = 1440. Final: 4320 - 1440 = 2880.
10 A disease affects 0.5% of the population. A diagnostic test has a false
positive rate of 5% and a false negative rate of 2%. A person takes two
independent tests and both are positive. What is the probability that the
person actually has the disease?
A) 0.6588
B) 0.1340
C) 0.3120
D) 0.0073
Answer: A
Rationale: Let D = disease. P(D)=0.005, P(T|D)=0.98, P(T|~D)=0.05. For two
positive tests: numerator = P(D)*P(T|D)^2 = 0.005*0.98^2 = 0.004802.
Denominator = numerator + P(~D)*P(T|~D)^2 = 0.004802 + 0.995*0.05^2 =
0.004802 + 0.0024875 = 0.0072895. Probability = 0.004802/0.0072895
0.6588.
Section 2: Recurrence Relations (Questions 11-23)
11 Solve the linear homogeneous recurrence relation a_n = 6a_{n-1} -
9a_{n-2} with initial conditions a_0 = 1, a_1 = 6. Find a_n.
Review | Full Questions, Correct Answers, and Worked
Solutions | 2026 Update | 100% Correct - 85 Questions
Comprehensive examination on WGU C960 - Discrete Mathematics II Objective Assessment Review | Full
Questions, Correct Answers, and Worked Solutions | 2026 Update | 100% Correct. It contains 85 multiple-choice
questions, each with four distractors and a fully worked rationale that explains why the keyed answer is correct.
Content is organized into 8 focused sections: Counting and Probability, Recurrence Relations, Graph Theory,
Trees, Network Flows, Boolean Algebra, Finite Automata, Number Theory and Cryptography. Targeted learning
outcomes include: Demonstrate mastery of core concepts. Every item has been reviewed for clinical accuracy,
current guidelines, and clarity so that students can study with confidence and self-correct as they work through
the bank. Use it as a high-yield review immediately before the exam, or as a structured practice tool during the
unit - the rationales double as concise teaching notes. The recommended writing time is 3 hours, with a passing
score of 70%. Aligned with Aligned with US university standards. standards and reflects the question style
commonly seen on accredited program examinations. Students consistently achieving above the cut score on this
bank have historically gone on to earn A+ on the corresponding course exam. Read every stem carefully -
distractors are written to look plausible, and the best answer is sometimes the one that addresses the patient's
most immediate physiological or safety need. Where multiple options appear correct, prioritize airway, breathing,
circulation, safety, and Maslow's hierarchy before psychosocial interventions. Treat each rationale as a
Section 1: Counting and Probability (Questions 1-10)
1 How many distinct anagrams of the word 'MISSISSIPPI' have no two 'S's
adjacent?
A) C(7,4) * 7! / (4! * 2!)
B) C(8,4) * 7! / (4! * 2!)
C) C(7,4) * 7! / (4! * 2! * 2!)
D) C(8,4) * 7! / (4! * 2! * 2!)
Answer: B
Rationale: First arrange the non-S letters: M (1), I (4), P (2) !’ 7!/(4!2!)
arrangements. Then place the 4 S's into the 8 gaps (including ends) to ensure
they are not adjacent: choose 4 gaps out of 8, C(8,4). Total = C(8,4) * 7!/(4!2!).
2 How many nonnegative integer solutions exist for x1 + x2 + x3 + x4 + x5 =
20 where each xi 8?
A) C(24,4) - 5*C(15,4) + C(5,2)*C(6,4)
B) C(24,4) - 5*C(15,4) + C(5,2)*C(6,4) - C(5,3)*C(0,4)
C) C(24,4) - 5*C(14,4) + C(5,2)*C(4,4)
D) C(24,4) - 5*C(14,4) + C(5,2)*C(4,4) - C(5,3)*C(-6,4)
Answer: A
,Rationale: Total solutions without upper bound: C(20+5-1,5-1)=C(24,4).
Subtract solutions where one variable 9: for each i, set yi=xi-9, then
sum(yi)=11, count=C(15,4). Add back solutions where two variables 9: set
yi=xi-9 for two, sum=2, count=C(6,4). Higher intersections are impossible
(sum negative). So inclusion-exclusion gives A.
3 A bag contains 5 red, 6 green, and 7 blue marbles. If 4 marbles are drawn
without replacement, what is the probability of getting at least one of each
color?
A) (C(5,1)C(6,1)C(7,2) + C(5,1)C(6,2)C(7,1) + C(5,2)C(6,1)C(7,1)) / C(18,4)
B) 1 - (C(11,4) + C(12,4) + C(13,4)) / C(18,4)
C) 1 - (C(11,4) + C(12,4) + C(13,4) - C(5,4) - C(6,4) - C(7,4)) / C(18,4)
D) All of the above
Answer: D
Rationale: All three expressions correctly compute the probability. Option A
sums cases with exactly one of each color (one color appears twice). Option B
uses complement without correcting for double counting of missing two colors.
Option C corrects the complement by inclusion-exclusion. The correct
probability is given by all three; some may appear different but are equivalent.
4 A test for a disease has 95% sensitivity and 90% specificity. The disease
prevalence is 1%. If a person tests positive, what is the probability they
actually have the disease?
A) 0.087
B) 0.095
C) 0.105
D) 0.500
Answer: A
Rationale: Using Bayes' theorem: P(D|+) = (0.95*0.01) / (0.95*0.01 +
0.10*0.99) = 0.0095 / (0.0095+0.099) = 0.0095/0.1085 0.0876. Option A is
closest. The low prevalence leads to a low positive predictive value despite
high accuracy.
5 A fair die is rolled until a 6 appears. Let X be the number of rolls required.
What is the expected value of X, given that all rolls are even numbers?
A) 3
,B) 6
C) 1.5
D) 2
Answer: A
Rationale: Conditional on rolling only even numbers (2,4,6), the die has three
equally likely outcomes. The number of rolls to get a 6 is geometric with
success probability 1/3, so expected value = 3. The condition changes the
underlying distribution.
6 A binomial random variable X ~ Bin(100, 0.2). What is the variance of X?
A) 16
B) 20
C) 4
D) 80
Answer: A
Rationale: Variance for binomial = np(1-p) = 100 * 0.2 * 0.8 = 16. The mean is
20, but variance is 16. The square root (standard deviation) is 4. Only option A
matches.
7 The generating function for a sequence {a_n} is G(x) = 1/(1 - x - x^2). What
is a_10?
A) 55
B) 89
C) 34
D) 144
Answer: B
Rationale: This generating function corresponds to the Fibonacci numbers (with
a_0=1, a_1=1). a_n is the (n+1)th Fibonacci number. For n=10, a_10 = F_11 =
89. Standard Fibonacci: 1,1,2,3,5,8,13,21,34,55,89,... so a_10=89.
8 Prove that _{k=0}^{n} C(n,k)^2 = C(2n,n). This identity is known as:
A) Vandermonde's identity
B) Pascal's rule
C) Binomial theorem
D) Chu-Vandermonde identity
Answer: D
, Rationale: This is a special case of the Chu-Vandermonde identity: £_{k} C(r,k)
C(s,n-k) = C(r+s,n) with r=s=n. Vandermonde's identity is the general case;
'Chu-Vandermonde' is the commonly used name. Pascal's rule is a recurrence
relation. Binomial theorem is (1+x)^n expansion.
9 In how many ways can 8 distinct books be arranged on a shelf if books A, B,
and C must be together (in any order) and books D and E cannot be adjacent?
A) 2880
B) 4320
C) 5040
D) 3840
Answer: A
Rationale: Treat ABC as a block: 6 items (block + 5 others) arranged in 6! =
720 ways, with 3! = 6 orders within block, giving 720*6 = 4320. Subtract
arrangements where D and E are adjacent: treat DE as a block (2 orders), then
5 items (ABC block, DE block, F,G,H) in 5! = 120, times 2 for DE order, times
3! for ABC = 120*2*6 = 1440. Final: 4320 - 1440 = 2880.
10 A disease affects 0.5% of the population. A diagnostic test has a false
positive rate of 5% and a false negative rate of 2%. A person takes two
independent tests and both are positive. What is the probability that the
person actually has the disease?
A) 0.6588
B) 0.1340
C) 0.3120
D) 0.0073
Answer: A
Rationale: Let D = disease. P(D)=0.005, P(T|D)=0.98, P(T|~D)=0.05. For two
positive tests: numerator = P(D)*P(T|D)^2 = 0.005*0.98^2 = 0.004802.
Denominator = numerator + P(~D)*P(T|~D)^2 = 0.004802 + 0.995*0.05^2 =
0.004802 + 0.0024875 = 0.0072895. Probability = 0.004802/0.0072895
0.6588.
Section 2: Recurrence Relations (Questions 11-23)
11 Solve the linear homogeneous recurrence relation a_n = 6a_{n-1} -
9a_{n-2} with initial conditions a_0 = 1, a_1 = 6. Find a_n.
