Probability: questions with answers

The test behaves identically in the two towns, but the answer is a share of the positive results, and where the disease is ten times as common the true positives make up a far larger share of them, so the probability is larger in town Q. That is exactly why the prior cannot be dropped. Saying the two are equal to the detection rate reports the test's reliability instead of the answer to the question; saying they equal the two prevalences treats a positive result as carrying no information at all; and a denominator counting only the people who have the disease would leave the false positives out of the very total they belong to.

"How likely is it that this faulty bulb came from the first supplier" conditions on the evidence - the bulb being faulty - and asks about the cause, so it is $P(\text{first supplier} \mid \text{faulty})$. The fault rate of the first supplier is the same two events conditioned the other way round, and the two are equal only by coincidence; the share of the store that comes from the first supplier is where the calculation starts, not where it ends; and the probability that a bulb is both faulty and from the first supplier is the numerator of the quotient, not the quotient itself.

$F = \{1, 2, 3\}$ and $E = \{1, 2, 3, 4\}$, so $E \cap F = \{1, 2, 3\} = F$ and $P(E \mid F) = \dfrac{3}{3} = 1$. Once the number is known to be at most $3$ it is certainly at most $4$, so the answer is a boundary value, not a fraction. Dividing $3$ by the full $6$ ignores the condition, and dividing $n(E) = 4$ by $n(F) = 3$ counts a favourable outcome, $4$, that the condition has excluded.

$E = \{HHH, HHT, HTH, THH\}$ and $F$ is every outcome except $HHH$, so $n(F) = 7$ and $E \cap F = \{HHT, HTH, THH\}$ with $3$ outcomes, giving $\dfrac{3}{7}$. Reading both quantifiers as "exactly two" makes $E$ and $F$ the same three-outcome event and produces $1$, which is why that reading feels safe and is not. Dividing by $8$ keeps $HHH$, which the condition has ruled out.

Since $A \subset B$, the intersection $A \cap B$ is $A$ itself, so $P(A \mid B) = \dfrac{P(A)}{P(B)}$. Because $P(B) \leq 1$, dividing by it cannot make the value smaller, so $P(A \mid B) \geq P(A)$, with equality only when $P(B) = 1$. Conditioning on a set that already contains $A$ concentrates the space, it does not dilute it.

Both conditional probabilities are written, so both $P(A)$ and $P(B)$ must be nonzero. The two quotients share the numerator $P(A \cap B)$, so $\dfrac{P(A \cap B)}{P(B)} = \dfrac{P(A \cap B)}{P(A)}$ forces $P(A) = P(B)$. It does not force the events to be equal: two different events can have equal probabilities. It also says nothing about $P(A \cap B) = P(A)P(B)$, which is independence.

$P(B \mid A) = 1$ means $\dfrac{P(A \cap B)}{P(A)} = 1$, so $P(A \cap B) = P(A)$: inside $A$ the event $B$ is certain, which says every outcome of $A$ lies in $B$. The containment therefore runs from the conditioning event into the target, $A \subset B$, and not the other way round. It does not make $P(B) = 1$, and the complement statement about $A'$ is a different reduced space entirely.

$P(\text{both black}) = P(\text{first black})\,P(\text{second black} \mid \text{first black})$. The first factor is $\dfrac{ 7 }{ 18 }$; once a black ball is out, $6$ black balls remain among $17$, so the second factor is $\dfrac{ 6 }{ 17 }$. Repeating the first factor would be correct only with replacement, and reducing the total while leaving the black count unchanged forgets that the ball removed was itself black.

Example 13 proves that $P(E \cap F') = P(E) - P(E)P(F) = P(E)P(F')$, so $E$ and $F'$ are independent; the same argument applied twice shows that $E'$ and $F$ are independent and that $E'$ and $F'$ are independent. Independence therefore survives complementing either event, or both, which is exactly what makes the "at least one" formula legitimate. A product of the events themselves subtracted from $1$ is the complement of "both occur", not of "neither occurs".

Independence is $P(A \cap B) = P(A)P(B)$, and applying it to the complements gives $P(A' \cap B') = [1 - P(A)][1 - P(B)]$, which is an equivalent test. $P(A \cup B) = P(A) + P(B)$ is the condition for the events to be mutually exclusive, which for events of positive probability actually rules independence out. Equal probabilities and probabilities summing to $1$ say nothing about the intersection at all.

Mutually exclusive events share no outcome, so $P(A \cap B) = 0$, while $P(A)\,P(B)$ is a product of two positive numbers and is therefore positive. The two are unequal, so the product rule fails and the events are dependent - in fact disjointness is the strongest possible dependence, since knowing that $A$ has occurred makes $B$ impossible. Independent events with positive probabilities do overlap, and $P(A \cap B) = 0$ satisfies the product rule only when one of the probabilities is $0$.

The partition of a sample space is not unique: the chapter itself gives the pair $E$ and $E'$, and the four-region family built from $E$, $F$ and their complements, as two different partitions of the same space. The definition asks that the parts be pairwise disjoint, exhaustive and of positive probability - it never asks that they be equally likely, it never allows a part of probability zero, and disjointness on its own is not enough.

A strike and no strike form a partition, so by the theorem of total probability $P(\text{on time}) = P(\text{strike})\,P(\text{on time} \mid \text{strike}) + P(\text{no strike})\,P(\text{on time} \mid \text{no strike})$: multiply along each branch of the tree and add the two products. Each branch contributes in proportion to how often it is taken, so averaging the two conditional probabilities is right only when the two branches are equally likely, which they are not here. Pairing each weight with the other branch's conditional probability uses the same four numbers rearranged and looks equally tidy, and keeping only the strike branch throws away every way of finishing on time without a strike.

The theorem of total probability multiplies each branch probability by the conditional probability BELONGING to that branch and adds the products, so the next line pairs $\dfrac{3}{12}$ with $\dfrac{3}{10}$, $\dfrac{4}{12}$ with $\dfrac{5}{10}$ and $\dfrac{5}{12}$ with $\dfrac{8}{10}$. Pairing them in any other order uses the same six numbers and looks equally tidy; using the raw ratio numbers $3$, $4$ and $5$ as the weights abandons weights that sum to $1$ and gives an answer far bigger than $1$; and stopping after the first product drops the two branches on which Q or R is selected.

Several outcomes can be sent to the same value: the outcomes with exactly $1$ heads differ in WHERE the heads fall, and there are $5$ such arrangements. Answering $1$ assumes each value comes from a single outcome, which would make $X$ a relabelling of the sample space rather than a measurement made on it. The fraction $\dfrac{ 5 }{ 32 }$ is the probability that $X = 1$, not a number of outcomes, and $6$ counts the values $X$ can take rather than the outcomes that give this one.