What is Conditional probability?

Example 32.6 (Screening test)

A disease affects $1\%$ of a population. A test detects it with probability $0.99$ (sensitivity) and gives a false positive with probability $0.05$. Given a positive test, the probability of actually having the disease is

Despite the accurate test, five positives out of six are false — because the disease is rare. Confusing $\pcond{A}{B}$ with $\pcond{B}{A}$ is the prosecutor’s fallacy.

Example 21.12 (Conditioning can create uniformity)

Roll two fair dice and condition on the sum being $7$: for each $k \in \intint16$,

given a sum of $7$, the first die is exactly uniform — $7$ is the only total compatible with every face, so the conditioning erases all information about $X$. Any other total skews the law (given $S = 4$, the first die is uniform on $\{1, 2, 3\}$ only). Computing a conditional law means renormalizing the joint weights along the conditioning event, nothing more.

Example 21.13 (The second draw is as good as the first)

An urn holds $3$ white and $2$ black balls; draw two without replacement. Everyone agrees $\P(W_1) = \frac35$; what is $\P(W_2)$? Total probability along the first draw:

exactly $\P(W_1)$. No computation was needed: by symmetry, every ball is equally likely to be the second one drawn, so the second draw — unconditionally — has the same law as the first. Conditioning on the first result changes the odds; not knowing it does not. This exchangeability argument returns in the next chapter for sampling without replacement, where it gives the hypergeometric mean $np$ with no binomial identities at all.

Example 21.15 (The birthday collision, by the chain rule)

With $n$ people whose birthdays are independent and uniform over $365$ days, let $D_n = {}$“all $n$ birthdays differ”. Conditioning person by person (chain rule):

each new person having to avoid the $k$ days already taken. For $n = 23$: $\P(D_{23}) \approx 0.493$ — a shared birthday is already more likely than not. The heuristic that explains the smallness of $23$: taking logarithms, $-\ln \P(D_n) \approx \sum_{k<n}\frac k{365} = \frac{\binom n2}{365}$, and $\binom{23}2 = 253$ gives $253/365 \approx 0.693 \approx \ln 2$. What matters is the number of pairs, which grows quadratically: collision problems live on the scale $n \sim \sqrt{365}$, not $n \sim 365$ — the birthday paradox is a square root in disguise.