This Exam P sample reference tests Bayes' Theorem. Normalizing the three proportional category probabilities gives priors 0.5, 0.4, and 0.1. Their zero-count contributions are 0.05, 0.08, and 0.04, so the requested posterior is 0.04/0.17=0.235294 and choice C.
These notes identify the calculation error associated with each wrong letter when that error is reproducible.
AThis is the joint contribution P(C and zero claims)=0.1(0.4)=0.04. It has not been divided by the total zero-claim probability.
BThis is the evidence probability P(zero claims)=0.05+0.08+0.04=0.17, which is the denominator rather than the requested posterior.
DThis is the posterior probability of category A, 0.05/0.17=0.294118. It normalizes the wrong category contribution.
EThis is the posterior probability of category B, 0.08/0.17=0.470588. It likewise uses the wrong numerator.
Original practice · fully worked
Original variant: mode inference from a continuous score
A signal is generated in mode R, S, or T with prior probabilities 0.45, 0.35, and 0.20. Conditional on the mode, its score Z lies between 0 and 1 with densities f_R(z)=2z, f_S(z)=2(1-z), and f_T(z)=1. A score of 0.80 is observed. Calculate the posterior probability that the signal used mode T.
A 0.132
B 0.189
C 0.200
D 0.450
E 0.679
Variant answer in brief
At the observed score, the prior-weighted density contributions of modes R, S, and T are 0.72, 0.14, and 0.20. Normalizing T's contribution gives 0.20/1.06=0.188679, so choice B.
Setup
Setup
Evaluate each conditional density at the observed score.
fR(0.8)=1.6,fS(0.8)=0.4,fT(0.8)=1
Model
Model
For a continuous observation, Bayes' theorem uses likelihood densities. Multiply each density by its mode prior.
wR=0.45(1.6)=0.72
wS=0.35(0.4)=0.14,wT=0.20(1)=0.20
Compute
Compute
Normalize mode T's contribution by the mixture density at the observation.
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