Practice: Conditional Probability, Total Probability and Bayes' Rule

(a) The table, both directions. Row totals: studied $45+15=60$, did not $10+30=40$. Column totals: passed $45+10=55$, failed $15+30=45$. Grand total 100 ✓.

Conditioning on studying discards the other row and renormalises within the 60 that remain:

Conditioning the other way divides by the column of 55 instead:

Same numerator 45, different denominators 60 and 55. The first divides by a row total, the second by a column total, and that is the whole of the asymmetry: conditioning on an event means dividing by that event's probability, so changing which event is conditioned on changes the denominator. Dividing by the grand total of 100 instead would give the joint probability $0.45$, which is a third quantity again.

(b) Independence. It would require $P(\text{studied}\cap \text{passed})=P(\text{studied})P(\text{passed})$:

Not equal, so they are dependent ✓. The conditionals say the same thing more vividly: $P(\text{passed}\mid \text{studied})=0.75$ against $P(\text{passed}\mid \text{did not})=\frac{10}{40}=0.25$, a threefold difference, where independence would have required both to equal the marginal 0.55.

Check by total probability: studying and not studying partition the students, so

matching the column total exactly.

(c) The positive test. Base rate first: $P(D)=0.01$. Likelihoods: $P(+\mid D)=0.99$, $P(+\mid \text{not }D)=1-0.95=0.05$. The question asks for $P(D\mid +)$, the reverse of the 99% given.

By natural frequencies, over 10,000 people:

1% of 10,000 is 100 diseased, and 99% of them test positive: 99. Of the 9,900 healthy, 5% test positive anyway: 495. Conditioning on a positive result restricts to the 594 in that column:

By Bayes' rule. The denominator comes from total probability:

Same answer exactly, since they are the same computation, with the table making the denominator visible as a count.

Why so far from 99%. Two reasons, both structural. First, 99% was the wrong conditional: $P(+\mid D)$ divides by the 100 diseased, $P(D\mid +)$ by the 594 positives. Second, the base rate dominates: there are 99 times more healthy people, so a mere 5% false-positive rate among them yields 495 false alarms against 99 true ones, five to one. The test was not useless: it raised the probability from 1% to about 17%, a seventeenfold increase.

(d) The same test, a different population. With a 50% base rate the likelihoods are unchanged but the weights are not:

about 95% rather than 17%. Identical test, identical sensitivity and specificity, different population, and the posterior moves by a factor of more than five.

So a posterior is not a property of the test. It is a property of the test and the population it was applied to, which is why it must be reported with the base rate it assumes. A screening figure quoted in a symptomatic clinic, or the reverse, is not a conservative approximation; it is a different number answering a different question.