Derive: Probability as Reasoning

By the definition of conditional probability, $Q(A)=P(A\text{ and }B)/P(B)$. For the whole sample space, $Q(\Omega )=P(\Omega \text{ and }B)/P(B)=P(B)/P(B)=1$. Now let $A_1$ and $A_2$ be disjoint. The events $A_1\cap B$ and $A_2\cap B$ are disjoint as well, because any outcome in both would lie in $A_1$ and in $A_2$. The sum rule for $P$ therefore gives $$P((A_1\cup A_2)\cap B)=P(A_1\cap B)+P(A_2\cap B).$$ Divide both sides by $P(B)$: $$Q(A_1\cup A_2)=Q(A_1)+Q(A_2).$$ The complement rule follows by taking $A_2$ to be the complement of $A_1$: the two are disjoint and their union is $\Omega$, so $Q(A_1)+Q(A_1^c)=1$. Conditioning therefore produces a new belief assignment that obeys the same rules as the old one. This is why the whole chapter can condition first and reason afterwards: every sum-rule identity proved before conditioning survives inside the condition. In the email table of section 11.3, for example, $P(\text{spam}\mid \text{free})+P(\text{not spam}\mid \text{free})=30/46+16/46=1$. The result does not say that $Q$ agrees with $P$ on any event; it says that $Q$ is internally consistent in the same way.

The four ordered outcomes are HH, HT, TH and TT, each with belief $1/4$. The event $A$ contains HH and HT, so $P(A)=1/2$. The event $B$ contains HH and TH, so $P(B)=1/2$. The event $C$ contains HH and TT, so $P(C)=1/2$. For the pair $A$ and $B$, the overlap is HH alone, so $P(A\text{ and }B)=1/4$, which equals $P(A)P(B)=1/4$. For $A$ and $C$ the overlap is also HH, and for $B$ and $C$ it is also HH, so both pairs satisfy the product form. Every pair is independent. The triple overlap is the single outcome HH, so $P(A\text{ and }B\text{ and }C)=1/4$. The product of the three separate beliefs is $1/2\times 1/2\times 1/2=1/8$. The two numbers differ, so the three events are not jointly independent even though every pair is. Knowing $A$ and $B$ together tells you $C$ with certainty: if both coins show heads, they agree. The product form for two events therefore does not extend to three by multiplication alone, and an independence assumption stated pairwise is weaker than one stated for the whole collection.

Bayes' rule gives $$P(\text{condition}\mid \text{positive})=\frac{0.90p}{0.90p+0.05(1-p)}.$$ Set this equal to $1/2$: $$2(0.90p)=0.90p+0.05(1-p).$$ Subtract $0.90p$ from both sides: $$0.90p=0.05(1-p)=0.05-0.05p.$$ Collect the $p$ terms: $0.95p=0.05$, so $p=0.05/0.95=1/19=0.0526$. At that prevalence the two routes to a positive result are equally common: the true positives contribute $0.90/19$, and the false positives contribute $0.05\times 18/19$, and the two are equal. The posterior is then exactly $1/2$: the positive result moves the odds from 1 to 18 against up to 1 to 1. Below that prevalence the false positives outnumber the true positives; above it the true positives dominate. The threshold belongs to this test's two error rates, and a test with a smaller false-positive rate has a lower threshold.