Practise: Probability as Reasoning

The both cell is 0.5. The parcel-only cell is $0.6-0.5=0.1$, and the invoice-only cell is $0.7-0.5=0.2$. The three cells that contain at least one event total $0.5+0.1+0.2=0.8$, so the neither cell is $1-0.8=0.2$. The sum rule gives the same number by the other route: $0.6+0.7-0.5=0.8$. The table is now complete, its four cells add to one, and its row and column totals return the three original beliefs. The subtraction of the overlap is the whole correction: adding 0.6 and 0.7 alone would count the parcel and invoice arriving together twice and return 1.3, which is not a belief at all.

The ordered outcomes are HHH, HHT, HTH, HTT, THH, THT, TTH and TTT, which is $2^3=8$ outcomes. Exactly two heads occurs in HHT, HTH and THH, so three outcomes lie inside the event. The belief is $3/8=0.375$. The order matters: HHT and HTH are different outcomes, and collapsing them into one result labelled "two heads" would leave a space of four outcomes that are not equally likely. The counting rule requires the finest grain the experiment distinguishes, which here is the position of the single tail.

The condition narrows the folder to the $40+32=72$ messages that contain offer. Inside that narrowed space, 40 are spam, so the belief is $40/72=0.556$. The reverse conditional belief is $40/80=0.5$, and it answers a different question: how many spam messages contain offer. The two numbers differ because the denominator of the first is the offer column (72) and the denominator of the second is the spam row (80). The remaining 288 ordinary messages without offer and the 40 spam messages without offer drop out of the first calculation entirely, and they cannot change its answer.

The clinic holds $50,000/500=100$ people with the condition, and the test finds $0.95\times 100=95$ of them. The remaining 49,900 people do not have the condition, and $0.02\times 49,900=998$ of them receive a false positive. Positive results therefore number $95+998=1093$. The share belonging to people with the condition is $95/1093=0.087$. The sensitivity alone says 0.95, and it is the wrong number for this question: it counts sick people who test positive, while the clinic needs positive people who are sick. The healthy majority supplies 998 of the 1093 positives, so it controls the answer even though each healthy person is unlikely to test positive.

The product form requires $P(A\text{ and }B)=P(A)P(B)=0.4\times 0.5=0.2$. The observed joint belief is 0.15, which is below 0.2, so the two events are not independent. The same conclusion appears in the conditional belief: $P(A\mid B)=0.15/0.5=0.3$, which is below the original $P(A)=0.4$. Learning $B$ therefore lowers the belief in $A$ rather than leaving it alone, and the definition of independence from section 11.5 fails. The mismatch is small in absolute terms and still decisive, because independence is an equality and not a rough agreement.

A dry hour has belief $0.85$. Under independence, the belief that all eight hours are dry is the product $0.85\times 0.85\times \cdots \times 0.85=0.85^8=0.272$. The same computation with twelve hours at 0.9 gives $0.9^{12}=0.282$, which is the number section 11.5 uses to show what the independence assumption would imply. The arithmetic is correct and the assumption is the fragile part: rain persists, so a wet hour raises the belief in a wet next hour, and the true belief in a dry day is not the product. The number therefore belongs to the independence model, not to the weather.

The announced rate predicts $0.40\times 250=100$ wet days. The record contains 92, so the observed frequency is $92/250=0.368$. The forecast ran 0.032 below its announced rate, which is 3.2 percentage points. A frequentist reads this as a measurement of the forecaster's long-run rate, slightly low; a Bayesian reads it as evidence about the forecaster's calibration, which is the match between announced probabilities and observed frequencies over many cases. Neither reading changes the arithmetic. The record covers 250 days from one forecaster, and it does not by itself establish the rate for a different forecaster, a different season or a different city.

Options a, c and d each describe an event inside a repeatable process: rolls of a die, a hundred rolls, or draws from a patient population. A frequentist can imagine the process repeated and read the probability as a long-run share. The particular number in option b is either prime or not, and no repetition of that same number is available, so the frequentist reading has no long-run frequency to offer for the statement. A frequentist can still estimate the density of primes in that range and quote that estimate, because the repetition is over the numbers in the range. A Bayesian assigns a belief to the statement directly and revises it when a primality test returns evidence. Section 11.6 uses this contrast to show that the readings differ over which events may carry a probability at all, not over the arithmetic once a probability is assigned.