This Exam P sample reference tests Conditional Probability. Conditioning must retain the driver's fixed risk class. Mixing the two class-specific two-year probabilities gives a joint numerator of 0.00369, while the current-year accident probability is 0.057; their ratio is 0.06474, which rounds to 0.065 and choice D.
How to solve this Conditional Probability question
Setup
Setup
Let H and L denote the two fixed risk classes, and let T and N denote accidents in the current and following years. Compute the current-year accident probability by mixing over class.
Pr(T)=0.10(0.12)+0.90(0.05)=0.057
Model
Model
The two years are independent only after conditioning on class. Therefore, square each class-specific annual probability before mixing the two-year joint event.
Pr(T∩N)=0.10(0.12)2+0.90(0.05)2
Compute
Compute
Evaluate the joint numerator and normalize it by the observed current-year event.
Pr(T∩N)=0.00144+0.00225=0.00369
Pr(N∣T)=0.0570.00369=0.0647368421…
Answer
Answer
The conditional next-year probability rounds to 0.065.
0.065(D)
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AThis is the joint two-year probability 0.00369, rounded to 0.004. It has not been divided by the conditioning probability 0.057.
BThis is approximately (0.12)²=0.0144, the probability of two accidents conditional on the high-risk class alone. It ignores both the class mixture and the stated conditioning.
CThis is the unconditional annual mixture 0.10(0.12)+0.90(0.05)=0.057. Observing a current accident raises the posterior weight on the higher-risk class.
EA mixture of 0.12 and 0.05 equal to 0.099 would require a high-risk posterior weight of 0.70. Bayes' rule gives only 0.012/0.057=4/19, so this choice over-updates the class probability.
Original practice · fully worked
Original variant: infer seed-lot type from split assay results
A sealed seed lot is vigorous with probability 0.25 and ordinary with probability 0.75. Conditional on its type, two germination assays are independent. Each assay is positive with probability 0.60 for a vigorous lot and 0.20 for an ordinary lot. Exactly one of the two assays is positive. Calculate the probability that the lot is vigorous.
A 0.1200
B 0.2500
C 0.3200
D 0.3333
E 0.4800
Variant answer in brief
The exactly-one-positive likelihood is 0.48 for a vigorous lot and 0.32 for an ordinary lot. Prior weighting gives joint masses 0.12 and 0.24, so the vigorous posterior is 0.12/(0.12+0.24)=1/3 and choice D.
Setup
Setup
Let E denote exactly one positive result. Compute its two class-conditional likelihoods from the two possible positive positions.
Pr(E∣V)=2(0.60)(0.40)=0.48
Pr(E∣O)=2(0.20)(0.80)=0.32
Model
Model
Multiply each likelihood by its lot-type prior to obtain the two joint weights for the observed assay pattern.
Pr(V∩E)=0.25(0.48)=0.12
Pr(O∩E)=0.75(0.32)=0.24
Compute
Compute
Normalize the vigorous-lot joint weight over both possible lot types.
Pr(V∣E)=0.12+0.240.12=31
Answer
Answer
The posterior probability that the lot is vigorous is approximately 0.3333.
The 3108-page Probability Proof Manual reorganizes 718 verified Exam P solutions by syllabus skill and adds formula proofs, error patterns, and original worked practice.
