WGU C955 Applied Probability and
Statistics 2025/2026 – Verified
Questions & Expert Rationales |
Updated Set
Question 1:
The probability of drawing an ace from a standard deck of 52 cards is:
A. 1/13 B. 4/52 C. 4/52 D. 1/4
Rationale: There are 4 aces in 52 cards, so P(ace) = favorable outcomes / total = 4/52 = 1/13;
this basic probability calculation illustrates the classical definition, useful for risk assessment in
decision-making scenarios like card games or sampling.
Question 2:
For independent events A and B, P(A and B) = :
A. P(A) + P(B) B. P(A) / P(B) C. P(A) × P(B) D. P(A|B) + P(B)
Rationale: Independence means occurrence of one doesn't affect the other, so joint probability
multiplies marginals; e.g., P(heads on coin 1 and even on die) = 0.5 × 0.5 = 0.25, foundational
for modeling uncorrelated business risks.
Question 3:
The complement rule states P(not A) = :
A. 1 - P(A) B. 1 - P(A) C. P(A) / (1 - P(A)) D. 1 + P(A)
Rationale: Total probability sums to 1, so P(A^c) = 1 - P(A); e.g., P(not default) = 1 - 0.05 =
0.95, simplifying calculations for binary outcomes in credit scoring.
,Question 4:
In a uniform distribution from 1 to 10, P(X = 5) = :
A. 0.1 B. 1/10 C. 1/10 D. 0.5
Rationale: Discrete uniform assigns equal probability 1/n; here n=10, so each integer has P=0.1,
modeling fair scenarios like random selection for quality assurance sampling.
Question 5:
Addition rule for mutually exclusive events: P(A or B) = :
A. P(A) × P(B) B. P(A) + P(B) C. P(A) + P(B) D. P(A|B)
Rationale: No overlap means union adds probabilities; e.g., P(even or ace) = 26/52 + 4/52 =
30/52, avoiding double-counting in event partitioning for market segmentation.
Question 6:
Conditional probability P(A|B) = :
A. P(A and B) / P(A) B. P(A and B) / P(B) C. P(B|A) × P(A) D. 1 - P(B)
Rationale: Given B, it's joint divided by marginal of conditioner; e.g., P(rain|clouds) = P(rain
and clouds)/P(clouds), essential for Bayesian updating in predictive analytics.
Question 7:
For two dice, P(sum = 7) = :
A. 1/6 B. 6/36 C. 6/36 D. 1/12
Rationale: 6 ways (1+6, 2+5, etc.) out of 36 outcomes; enumerates sample space for discrete
uniform, applied in simulation for operational risk modeling.
Question 8:
, Bayes' theorem formula: P(A|B) = :
A. [P(B|A) × P(A)] / P(B) B. [P(B|A) × P(A)] / P(B) C. P(A) + P(B|A) D. P(B) / P(A|B)
Rationale: Updates prior with likelihood over evidence; e.g., diagnostic tests, where
P(disease|positive) incorporates false positives for accurate medical/business diagnostics.
Question 9:
Expected value for a lottery (win $100 at 0.01, lose $1 at 0.99) = :
A. $0.01 B. $0.01 × $100 + 0.99 × (-$1) = -$0.99 C. $0.01 × $100 + 0.99 × (-$1) = -$0.99 D.
$99
Rationale: E[X] = Σ P_i × x_i; negative EV shows house edge, guiding investment decisions by
quantifying long-term outcomes.
Question 10:
Variance for Bernoulli trial p=0.5 = :
A. p B. p(1-p) C. p(1-p) D. p²
Rationale: Var = E[X²] - (E[X])² = p - p² = p(1-p); max at p=0.5, measures uncertainty in binary
models like conversion rates.
Question 11:
In binomial n=10, p=0.3, P(X=3) uses:
A. Poisson approximation B. C(10,3) (0.3)^3 (0.7)^7 C. C(10,3) (0.3)^3 (0.7)^7 D. Normal
Rationale: Binomial PMF for fixed trials/success probability; calculates defect probabilities in
manufacturing quality control.
Question 12:
Normal approximation to binomial valid when np and n(1-p) > :
Statistics 2025/2026 – Verified
Questions & Expert Rationales |
Updated Set
Question 1:
The probability of drawing an ace from a standard deck of 52 cards is:
A. 1/13 B. 4/52 C. 4/52 D. 1/4
Rationale: There are 4 aces in 52 cards, so P(ace) = favorable outcomes / total = 4/52 = 1/13;
this basic probability calculation illustrates the classical definition, useful for risk assessment in
decision-making scenarios like card games or sampling.
Question 2:
For independent events A and B, P(A and B) = :
A. P(A) + P(B) B. P(A) / P(B) C. P(A) × P(B) D. P(A|B) + P(B)
Rationale: Independence means occurrence of one doesn't affect the other, so joint probability
multiplies marginals; e.g., P(heads on coin 1 and even on die) = 0.5 × 0.5 = 0.25, foundational
for modeling uncorrelated business risks.
Question 3:
The complement rule states P(not A) = :
A. 1 - P(A) B. 1 - P(A) C. P(A) / (1 - P(A)) D. 1 + P(A)
Rationale: Total probability sums to 1, so P(A^c) = 1 - P(A); e.g., P(not default) = 1 - 0.05 =
0.95, simplifying calculations for binary outcomes in credit scoring.
,Question 4:
In a uniform distribution from 1 to 10, P(X = 5) = :
A. 0.1 B. 1/10 C. 1/10 D. 0.5
Rationale: Discrete uniform assigns equal probability 1/n; here n=10, so each integer has P=0.1,
modeling fair scenarios like random selection for quality assurance sampling.
Question 5:
Addition rule for mutually exclusive events: P(A or B) = :
A. P(A) × P(B) B. P(A) + P(B) C. P(A) + P(B) D. P(A|B)
Rationale: No overlap means union adds probabilities; e.g., P(even or ace) = 26/52 + 4/52 =
30/52, avoiding double-counting in event partitioning for market segmentation.
Question 6:
Conditional probability P(A|B) = :
A. P(A and B) / P(A) B. P(A and B) / P(B) C. P(B|A) × P(A) D. 1 - P(B)
Rationale: Given B, it's joint divided by marginal of conditioner; e.g., P(rain|clouds) = P(rain
and clouds)/P(clouds), essential for Bayesian updating in predictive analytics.
Question 7:
For two dice, P(sum = 7) = :
A. 1/6 B. 6/36 C. 6/36 D. 1/12
Rationale: 6 ways (1+6, 2+5, etc.) out of 36 outcomes; enumerates sample space for discrete
uniform, applied in simulation for operational risk modeling.
Question 8:
, Bayes' theorem formula: P(A|B) = :
A. [P(B|A) × P(A)] / P(B) B. [P(B|A) × P(A)] / P(B) C. P(A) + P(B|A) D. P(B) / P(A|B)
Rationale: Updates prior with likelihood over evidence; e.g., diagnostic tests, where
P(disease|positive) incorporates false positives for accurate medical/business diagnostics.
Question 9:
Expected value for a lottery (win $100 at 0.01, lose $1 at 0.99) = :
A. $0.01 B. $0.01 × $100 + 0.99 × (-$1) = -$0.99 C. $0.01 × $100 + 0.99 × (-$1) = -$0.99 D.
$99
Rationale: E[X] = Σ P_i × x_i; negative EV shows house edge, guiding investment decisions by
quantifying long-term outcomes.
Question 10:
Variance for Bernoulli trial p=0.5 = :
A. p B. p(1-p) C. p(1-p) D. p²
Rationale: Var = E[X²] - (E[X])² = p - p² = p(1-p); max at p=0.5, measures uncertainty in binary
models like conversion rates.
Question 11:
In binomial n=10, p=0.3, P(X=3) uses:
A. Poisson approximation B. C(10,3) (0.3)^3 (0.7)^7 C. C(10,3) (0.3)^3 (0.7)^7 D. Normal
Rationale: Binomial PMF for fixed trials/success probability; calculates defect probabilities in
manufacturing quality control.
Question 12:
Normal approximation to binomial valid when np and n(1-p) > :
