1The idea in one paragraph

Most real information comes the wrong way round. A test manufacturer tells you how often the test is positive for people who have the condition, P(positive | condition). A patient wants to know how likely they are to have the condition given that the test was positive, P(condition | positive). These are different numbers, sometimes wildly different. Bayes' theorem converts one into the other. It uses three ingredients: how common each possible cause is before any evidence, P(B); how likely the evidence is under each cause, P(A | B); and the total probability of the evidence, which comes from adding up every route to it. The answer is always the same shape: the route through the cause you care about, divided by all the routes that produce what you saw.

2From the definition to the theorem

Start from the definition of conditional probability, aimed the way you want to go:

Now replace the top and the bottom with things you know. The top comes from the alternate form in 4.11: P(A ∩ B) = P(B)P(A | B). The bottom is the total probability of A. A happens either with B or with B′, and those two routes cannot overlap, so

Put both into the definition:

P(B | A) = P(B)P(A | B) ÷ [P(B)P(A | B) + P(B′)P(A | B′)] · Bayes' theorem, from the formula booklet.

Figure 1 shows where each piece sits on a tree. The tree is drawn with B first, so every quantity on the right-hand side of the formula is on a branch. The numerator is the product along the teal path, B then A. The denominator is the sum of the two paths that end in A, teal and amber. Bayes' theorem is nothing more than the "path over paths" recipe you used in 4.11, written in symbols.

So you always have a choice: quote the formula and substitute, or draw the tree and compute path over paths. Markers accept both. The formula is safer when the question gives you conditional probabilities in words and no tree; the tree is safer when you need several answers from the same situation.

3A screening test

Here is the context the guide names: medical studies. A screening test is used for a condition that affects 1% of the people screened. Among people who have the condition, the test is positive 95% of the time. Among people who do not, it is still positive 8% of the time (a false positive). A person tests positive. How likely is it that they have the condition?

Let D be "has the condition" and + be "tests positive". Then P(D) = 0.01, P(+ | D) = 0.95 and P(+ | D′) = 0.08.

A positive result means about an 11% chance of having the condition. Most people find that shocking; a test that is "95% accurate" sounds as though a positive result should mean 95%. Figure 2 shows why it does not, using natural frequencies: imagine 10 000 people. About 100 have the condition, and 95 of them test positive. Of the 9 900 without it, 8% is 792, and every one of those tests positive too. So 887 people test positive, and only 95 of them have the condition. The false positives swamp the true ones because there are so many more healthy people for the 8% to act on.

The 1% is called the prior probability, P(D) before any evidence; the answer 0.107 is the posterior probability, P(D | +) after it. You will not be asked to use these words, but they make the story easy to tell: the evidence raised the probability from 0.01 to 0.107, more than tenfold, yet not anywhere near certainty, because the prior was so small.

Figure 3 plots the same test used in groups where the condition is more or less common. The posterior rises steeply with the prior. In a group where 30% have the condition, a positive result means 0.836. The test did not change; the people did. That is why screening programmes aim tests at high-risk groups, and why a positive screening result is usually followed by a second, different test.

A negative result is a different question. P(D | −) uses the paths that end in a negative:

A negative result is very reassuring: about 5 in 10 000. Questions often ask for both, then ask you to comment. A good comment names both numbers: "the test is good at ruling the condition out, since P(D | −) is tiny, but poor at ruling it in, since P(D | +) is only 0.107."

4Updating twice: a second positive test

Suppose the person who tested positive is tested again with an independent repeat of the same test, and is positive again. Now the right starting point is not 1%. It is 0.107, what you know after the first test. Apply the theorem again with that as the prior. The assumption, which the question will state, is that the two results are independent for a person of known status.

Figure 4 shows the three stages. One positive raised 0.01 to 0.107; a second raised it to 0.588. You get the same 0.588 in one step by treating "positive twice" as the evidence, with P(++ | D) = 0.95² = 0.9025 and P(++ | D′) = 0.08² = 0.0064: then P(D | ++) = 0.009025 ÷ (0.009025 + 0.99 × 0.0064) = 0.588. Two routes, one answer, which is a good check.

5Bayes' theorem with three events

The guide asks for Bayes' theorem "for a maximum of three events". That means the possible causes are split into three: B₁, B₂ and B₃, which are mutually exclusive (only one can happen) and exhaustive (one of them must happen), so P(B₁) + P(B₂) + P(B₃) = 1. The theorem keeps its shape: one path on top, every path to A underneath.

P(Bᵢ | A) = P(Bᵢ)P(A | Bᵢ) ÷ [P(B₁)P(A | B₁) + P(B₂)P(A | B₂) + P(B₃)P(A | B₃)] · from the formula booklet.

A courier company sorts parcels at three centres: 45% go through the north centre (N), 35% through central (C) and 20% through south (S). The probability that a parcel is damaged (D) is 0.02 at N, 0.04 at C and 0.05 at S. A customer receives a damaged parcel. Which centre most probably handled it?

Figure 5 draws the tree with the three products.

Central is the most probable source (0.424), even though south has the worst damage rate. South handles too few parcels to dominate. North handles the most parcels, but its low damage rate keeps its share down. Bayes' theorem weighs both things at once: how much traffic each cause has, and how likely each is to produce the evidence.

Always check that the posteriors add to 1. Given the evidence, one of the three causes happened, so the three answers must sum to 1. It costs ten seconds and catches a wrong denominator.

Conditioning on the complement. "Given that a parcel arrived undamaged, find the probability that it went through the north centre" uses the D′ branches: P(D′) = 1 − 0.033 = 0.967, and P(N | D′) = 0.45 × 0.98 ÷ 0.967 = 0.441 ÷ 0.967 = 0.456 (3 s.f.). An undamaged parcel is slightly more likely than average to have come through N (0.456 against 0.45), because N rarely damages anything.

6Bayes on Paper 1: fractions

Without a calculator the numbers are chosen to be fractions with a common denominator in reach. Noor gets to school by bus with probability 1/2, by bike with probability 1/3, and on foot with probability 1/6. She is late with probability 1/5 by bus, 1/10 by bike and 1/4 on foot. One morning she is late. Find the probability that she cycled.

The bus paths give P(bus | L) = (1/10) ÷ (7/40) = 4/7 and walking gives 5/21; with 4/21 for the bike, the three sum to 12/21 + 4/21 + 5/21 = 1. Find a common denominator for P(L) before dividing: dividing by a sum of three unlike fractions is where Paper 1 answers go wrong.

7When the evidence tells you nothing: the link to independence

The guide links this subtopic to independent events. Suppose the evidence A is equally likely whatever the cause: P(A | B) = P(A | B′) = q. Then

The posterior equals the prior: seeing A does not change the probability of B at all. That is exactly the definition of independence from 4.11, P(B | A) = P(B). A test that is positive 20% of the time whether or not you have the condition is useless, and Bayes' theorem says so. The further apart P(A | B) and P(A | B′) are, the more the evidence moves the probability.

Working backwards. Because the theorem is an equation, a question can give the posterior and hide something else. For the screening test with P(D) = 0.01 and P(+ | D) = 0.95, how low must the false-positive rate x = P(+ | D′) be for a positive result to mean at least a 50% chance of having the condition?

The false-positive rate would need to fall from 8% to below about 0.96%. For a rare condition, a screening test needs a very low false-positive rate before a single positive result means much.