Reasoning Backward from Evidence

Problem
In a mathematical model, $0.1\%$ of a population has HIV. A test is positive for $90\%$ of those with HIV and for $1\%$ of those without it. Given a positive result, calculate the model probability that the person has HIV.

1. 1.Let $E$ mean has HIV and $A$ mean positive. The priors are $P(E)=0.001$, $P(E^{\prime })=0.999$. The likelihoods are $P(A\mid E)=0.9$, $P(A\mid E^{\prime })=0.01$.
2. 2.The true-positive contribution is $0.001(0.9)=0.0009$. The false-positive contribution is $0.999(0.01)=0.00999$. Thus $P(A)=0.01089$.
3. 3.The posterior is $P(E\mid A)=0.0009/0.01089=90/1089=10/121\approx 0.0826$. It is about $8.26\%$ within this stipulated model, not $90\%$.
4. 4.The $90\%$ describes positives among people with the condition. The posterior describes people with the condition among all positives. These are different groups. The numerical test properties here are assumptions of the problem.

Problem
Machines A, B, and C supply $25\%,35\%,40\%$ of bolts, with defect rates $5\%,4\%,2\%$. A randomly chosen bolt is defective. Find the probability that B made it.

1. 1.Let $E_A,E_B,E_C$ identify the producing machine and $D$ mean defective. Form the joint contributions: $0.25(0.05)=0.0125$, $0.35(0.04)=0.014$, and $0.40(0.02)=0.008$.
2. 2.Their sum is $P(D)=0.0345$. All three machine sources belong in the denominator.
3. 3.For machine B, $P(E_B\mid D)=0.014/0.0345=28/69\approx 0.4058$. Its posterior is about $40.58\%$, larger than its $35\%$ output share because its defect rate exceeds the overall rate.

Problem
A doctor uses train, bus, scooter, or another method with probabilities $3/10,1/5,1/10,2/5$. The corresponding lateness probabilities are $1/4,1/3,1/12,0$. Given a late arrival, find the probability of train travel.

1. 1.The transport choices form the partition. The four lateness contributions are $(3/10)(1/4)=3/40$, $(1/5)(1/3)=1/15$, $(1/10)(1/12)=1/120$, and $(2/5)(0)=0$.
2. 2.Add them using denominator $120$: $P(\text{late})=(9+8+1)/120=18/120=3/20$.
3. 3.The train posterior is $(3/40)/(3/20)=1/2$. Train lateness $P(\text{late}\mid \text{train})=1/4$ is the given likelihood; train given late is the requested posterior.

Problem
A fair die is rolled. A reporter truthfully answers the yes-or-no question “Was it a six?” with probability $3/4$, independently of the die result, and otherwise gives the opposite answer. Given a yes report, find the probability of an actual six.

1. 1.Let $S$ mean actual six and $A$ mean a yes report. Then $P(S)=1/6$, $P(S^{\prime })=5/6$, $P(A\mid S)=3/4$, and $P(A\mid S^{\prime })=1/4$.
2. 2.The contributions to a yes report are $(1/6)(3/4)=3/24$ and $(5/6)(1/4)=5/24$. Their total is $8/24$.
3. 3.The posterior is $(3/24)/(8/24)=3/8$. The explicit yes-or-no reporting assumption is what makes the second likelihood $1/4$. A model in which a liar chooses among five false die values would need different likelihoods.

Problem
Choose one of the four boxes in the table with equal probability and draw uniformly within it. Given a black ball, find the probability that box III was selected.

1. 1.The prior for every box is $1/4$. Their black likelihoods are $3/18=1/6$, $2/8=1/4$, $1/7$, and $4/13$.
2. 2.The equal prior factors cancel from the Bayes ratio, giving $P(\mathrm{III}\mid \text{black})=(1/7)/(1/6+1/4+1/7+4/13)$.
3. 3.With common denominator $1092$, the denominator sum is $(182+273+156+336)/1092=947/1092$. Thus the posterior is $156/947\approx 0.1647$.
4. 4.Using one black ball divided by all black balls would model a uniform choice from the pooled collection, which is a different experiment.

Problem
A machine setup is correct with probability $0.8$. An item is acceptable with probability $0.9$ under a correct setup and $0.4$ under an incorrect setup. Assume successive items are independent conditional on the setup. After two acceptable items, find the probability of a correct setup.

1. 1.Let $C$ mean correct setup and $A$ mean both selected items are acceptable. The priors are $0.8,0.2$.
2. 2.The conditional likelihoods are $P(A\mid C)=0.9^2=0.81$ and $P(A\mid C^{\prime })=0.4^2=0.16$. The joint contributions are $0.8(0.81)=0.648$ and $0.2(0.16)=0.032$.
3. 3.Divide the correct-setup contribution by their total: $P(C\mid A)=0.648/0.680=81/85\approx 0.9529$.
4. 4.The posterior is greater than the prior because two acceptable items are more likely under a correct setup. We assumed independence given the setup; unconditional independence of the two item outcomes was not required.