Two Tutorial 2 problems, from zero

Step 1Translate every sentence into a symbol before computing anything

Word problems in this chapter are lost or won at translation. Two events describe the truth: Pr = "impurity present", Ab = "impurity absent". Two describe one experiment's result: D = "detects", N = "does not detect". Now read the sheet one sentence at a time. The rule: whatever follows "if" or "given" goes to the right of the bar.

Step 2Build the likelihood of the data — three tests, not one

Bayes needs P(E | Pr): if the impurity is present, how likely is it that exactly two of three experiments detect it? Nothing in the sheet states this number. We build it from two ideas.

Idea 1 — independence lets you multiply. "Three separate experiments" means: given the truth, each experiment's result does not affect the others. For independent events, the probability that all of them happen is the product of their probabilities (§2.5). So the probability of any specific sequence of three results is a product of three per-test numbers. If the impurity is present, the sequence "detect, detect, no-detect" (write it DDN) has probability

Idea 2 — "exactly two detections" is several sequences, so list them. Three experiments, each D or N: by the product rule there are 2 × 2 × 2 = 8 possible sequences. Write them down in a fixed order (the systematic-listing habit from the prerequisite kit) and mark the ones with exactly two D's:

Three sequences have exactly two D's: DDN, DND, NDD (the N can sit in any one of the three positions — that "3" is C(3, 1) = C(3, 2) = 3, if you like the formula; listing is just as good here). Order doesn't change a product, so each of the three has the same probability. And they are mutually exclusive — the three tests produce one sequence, not two — so the addition rule for disjoint events applies:

Same construction in the other world. If the impurity is absent, a detection is a false alarm with probability 0.1 and a non-detection has probability 0.9:

Check before moving on. In each row of the figure, the eight probabilities add to 1: 0.512 + 3(0.128) + 3(0.032) + 0.008 = 1 and 0.001 + 3(0.009) + 3(0.081) + 0.729 = 1. If your factors are wrong, this check fails — it is the one-line audit for this kind of counting.

Step 3Now Bayes, exactly as in the notes

The data E can happen "via" two mutually exclusive routes — the impurity is present, or it is absent — and those two routes cover everything. That is the setting for total probability and Bayes (notes §2.4). Multiply along each route:

Bayes is the share of the data-world that belongs to the "present" route:

Answer: the posterior probability that the impurity is present is 0.1536/0.1698 = 0.9046, i.e. about 0.905 (to the quiz's five decimals: 0.90459).

Step 4The picture that turns Bayes into an area

The tree is the exam tool. This second picture is for understanding why the answer is so lopsided. Draw a square for "all batches". Split its width by the prior: 0.4 of it is "present", 0.6 is "absent". In each column, shade a height equal to the likelihood — the fraction of those batches that would show exactly two detections.

The absent column is wider (0.6 versus 0.4), but its shading is a sliver: two detections out of three is a rare event when the impurity is absent. In numbers, the data is 0.384/0.027 ≈ 14 times more likely under "present" than under "absent", and that factor of 14 overwhelms the prior's 0.6 : 0.4 lean towards "absent". That is all Bayes ever does: start from the prior areas, keep only the part consistent with the data, re-normalise.

Step 5Read the answer — don't just report it

Exams increasingly ask "interpret". Rerunning Steps 2–3 for every possible number of detections shows what the three experiments can and can't tell the engineer:

Two things to say in an exam answer: the evidence raised the probability from 0.40 to 0.905, and it did so because two-of-three is about fourteen times likelier with the impurity than without. Note also the one-detection row from your check box: a single detection would have lowered the engineer's belief to 0.21 — knowing the data can point either way is the mark of understanding Bayes rather than executing it.

Step 1Say the model out loud

Each person's birthday is one of 365 days, each with probability 1/365 ("equally likely"), and one person's birthday tells you nothing about another's ("randomly selected" — no twins in the sample). Dropping 29 February just makes the day count exactly 365 with all days equal. The outcome for k people is an ordered list of k days: (person 1's day, person 2's day, …). By the product rule there are 365 × 365 × ⋯ × 365 = 365^k such lists, and because every day is equally likely for every person, the lists are equally likely. That last sentence is what licenses "favourable over total" (§2.2 trap 2 on the notes page: check equal likelihood before you count).

Step 2"At least two share" is a mess; its complement is one clean case

Try to count "at least two share a birthday" directly and you meet cases: exactly one pair shares; two different pairs; three people on one day; a pair and a triple; … The complement, "all k birthdays are different", is a single case with a clean count. So we compute that and subtract:

This is the same move as Tutorial 1 Q10 and the parallel-system rule in §2.5: whenever the question says "at least one", price the "none" event and take the complement.

Step 3Two ways to price "all different" — and they agree

View 1, counting (§2.3). Build a favourable list one person at a time. Person 1 may have any of 365 days. Person 2 must avoid that day: 364 choices. Person 3 must avoid both: 363. … Person 10 must avoid nine taken days: 356. Product rule: 365 × 364 × ⋯ × 356 favourable lists (a permutation, P_10,365 in Devore's notation). Over 365¹⁰ equally likely lists:

View 2, a chain of conditional probabilities (§2.4). Pair each numerator factor with a denominator factor and the same fraction becomes a product of ten simple ones:

Read left to right, this is the multiplication rule applied nine times: the second person differs from the first with probability 364/365; given the first two differ, the third avoids both taken days with probability 363/365; and so on — the j-th new person must avoid j − 1 taken days, probability (365 − (j − 1))/365. Each factor lives in the world where everyone before already differs. Both views give the same number; View 2 is the one to compute with, because it never asks your calculator for 365¹⁰.

Answer (a): P(all ten different) ≈ 0.883, so P(at least two share) = 1 − 0.883 ≈ 0.117.

Step 4Part (b): keep multiplying until the complement crosses one half

"At least a 50–50 chance that two or more share" means 1 − P(all different) ≥ 0.5, i.e. the running product has fallen to 0.5 or below. There is no algebra that solves this for k in one line — the honest method is to continue the chain and watch:

Answer (b): at k = 22 the chance is 0.476 — not yet 50–50. At k = 23 it is 0.507. So the smallest such k is 23. (Whether you read "at least 50–50" as ≥ 0.5 or > 0.5 makes no difference here; write the reading you took anyway.)

Move the slider: the readout gives the exact running product for that k, and "Simulate one room" draws k random birthdays on a 365-day calendar strip and flags any day hit twice. Run twenty rooms at k = 23 and about half of them will show a shared birthday — the long-run relative frequency settling on the computed probability, exactly as in §2.2.

Step 5Why 23 is so small — the pairs intuition

Twenty-three feels far too few for a 365-day year because the mind pictures 23 chances of a match. But a shared birthday happens between a pair of people, and 23 people form C(23, 2) = 253 pairs. Each pair matches with probability 1/365; 253 pairs each with a 1-in-365 chance means you'd expect about 253/365 ≈ 0.69 matching pairs in the room. Once the expected number of matches is around 0.7, a room with at least one match being a coin flip is no longer surprising. (This is intuition, not a computation of the answer — pairs share people and are not independent, so adding pair probabilities overcounts; the chain in Steps 3–4 is the exact method.)