MATH 225N WEEK 4 STATISTICS QUIZ SOLUTIONS | Attempt
Score A+ | Chamberlain University | 2026/2027 Academic
Year | Pass Guaranteed - A+ Graded
[Section 1: Basic Probability Concepts (Questions 1-12)]
Q1. In a clinical trial of a new medication, the probability that a patient experiences
nausea (event A) is 0.42. What is the probability that a patient does NOT experience
nausea?
A. 0.42
B. 0.58
C. 1.42
D. 0.18
Correct Answer: B. 0.58 [CORRECT]
Rationale: By the complement rule, P(not A) = 1 − P(A) = 1 − 0.42 = 0.58. Option A is the
probability of nausea, not its complement. Option C incorrectly adds instead of
subtracting. Option D is 0.42 × 0.42, which has no probabilistic meaning here.
Q2. A hospital tracks two mutually exclusive complications after surgery: infection
(event B) with P(B) = 0.35 and bleeding (event C) with P(C) = 0.28. What is the
probability that a patient experiences either infection OR bleeding?
A. 0.098
B. 0.63
,C. 0.07
D. 0.35
Correct Answer: B. 0.63 [CORRECT]
Rationale: For mutually exclusive events, P(B or C) = P(B) + P(C) = 0.35 + 0.28 = 0.63.
Option A incorrectly multiplies (used for independent "and" problems). Option C
subtracts the probabilities. Option D is just P(B) alone.
Q3. In a patient population, the probability of having hypertension (event D) is 0.45, the
probability of having diabetes (event E) is 0.30, and the probability of having both is
0.12. What is the probability that a randomly selected patient has hypertension OR
diabetes?
A. 0.75
B. 0.63
C. 0.57
D. 0.135
Correct Answer: B. 0.63 [CORRECT]
Rationale: For non-mutually exclusive events, P(D or E) = P(D) + P(E) − P(D and E) = 0.45
+ 0.30 − 0.12 = 0.63. Option A forgets to subtract the intersection (classic trap). Option
C subtracts twice. Option D multiplies all three values incorrectly.
Q4. A diagnostic test for a disease has sensitivity P(F) = 0.60 and specificity P(G) = 0.40
(these represent independent events in this context). What is the probability that BOTH
sensitivity and specificity criteria are met?
A. 1.00
B. 0.24
C. 0.20
, D. 0.76
Correct Answer: B. 0.24 [CORRECT]
Rationale: For independent events, P(F and G) = P(F) × P(G) = 0.60 × 0.40 = 0.24. Option
A incorrectly adds probabilities. Option C averages them. Option D uses the
complement of the product.
Q5. In a study of 200 patients, 50 have condition H. Of those with condition H, 15 also
have condition I. What is the conditional probability P(I|H)?
A. 0.075
B. 0.30
C. 0.50
D. 0.15
Correct Answer: B. 0.30 [CORRECT]
Rationale: P(I|H) = P(H and I) / P(H) = (15/200) / (50/200) = 15/50 = 0.30. Option A
divides 15 by 200 directly. Option C is P(H). Option D is the joint probability P(H and I).
Q6. A screening test for a disease has prevalence 8%, sensitivity 94%, and specificity
88%. If a patient tests positive, what is the probability they actually have the disease
(positive predictive value)?
A. 0.94
B. 0.4052
C. 0.08
D. 0.88
Correct Answer: B. 0.4052 [CORRECT]
Score A+ | Chamberlain University | 2026/2027 Academic
Year | Pass Guaranteed - A+ Graded
[Section 1: Basic Probability Concepts (Questions 1-12)]
Q1. In a clinical trial of a new medication, the probability that a patient experiences
nausea (event A) is 0.42. What is the probability that a patient does NOT experience
nausea?
A. 0.42
B. 0.58
C. 1.42
D. 0.18
Correct Answer: B. 0.58 [CORRECT]
Rationale: By the complement rule, P(not A) = 1 − P(A) = 1 − 0.42 = 0.58. Option A is the
probability of nausea, not its complement. Option C incorrectly adds instead of
subtracting. Option D is 0.42 × 0.42, which has no probabilistic meaning here.
Q2. A hospital tracks two mutually exclusive complications after surgery: infection
(event B) with P(B) = 0.35 and bleeding (event C) with P(C) = 0.28. What is the
probability that a patient experiences either infection OR bleeding?
A. 0.098
B. 0.63
,C. 0.07
D. 0.35
Correct Answer: B. 0.63 [CORRECT]
Rationale: For mutually exclusive events, P(B or C) = P(B) + P(C) = 0.35 + 0.28 = 0.63.
Option A incorrectly multiplies (used for independent "and" problems). Option C
subtracts the probabilities. Option D is just P(B) alone.
Q3. In a patient population, the probability of having hypertension (event D) is 0.45, the
probability of having diabetes (event E) is 0.30, and the probability of having both is
0.12. What is the probability that a randomly selected patient has hypertension OR
diabetes?
A. 0.75
B. 0.63
C. 0.57
D. 0.135
Correct Answer: B. 0.63 [CORRECT]
Rationale: For non-mutually exclusive events, P(D or E) = P(D) + P(E) − P(D and E) = 0.45
+ 0.30 − 0.12 = 0.63. Option A forgets to subtract the intersection (classic trap). Option
C subtracts twice. Option D multiplies all three values incorrectly.
Q4. A diagnostic test for a disease has sensitivity P(F) = 0.60 and specificity P(G) = 0.40
(these represent independent events in this context). What is the probability that BOTH
sensitivity and specificity criteria are met?
A. 1.00
B. 0.24
C. 0.20
, D. 0.76
Correct Answer: B. 0.24 [CORRECT]
Rationale: For independent events, P(F and G) = P(F) × P(G) = 0.60 × 0.40 = 0.24. Option
A incorrectly adds probabilities. Option C averages them. Option D uses the
complement of the product.
Q5. In a study of 200 patients, 50 have condition H. Of those with condition H, 15 also
have condition I. What is the conditional probability P(I|H)?
A. 0.075
B. 0.30
C. 0.50
D. 0.15
Correct Answer: B. 0.30 [CORRECT]
Rationale: P(I|H) = P(H and I) / P(H) = (15/200) / (50/200) = 15/50 = 0.30. Option A
divides 15 by 200 directly. Option C is P(H). Option D is the joint probability P(H and I).
Q6. A screening test for a disease has prevalence 8%, sensitivity 94%, and specificity
88%. If a patient tests positive, what is the probability they actually have the disease
(positive predictive value)?
A. 0.94
B. 0.4052
C. 0.08
D. 0.88
Correct Answer: B. 0.4052 [CORRECT]
