1The idea in one paragraph

A probability always depends on what you already know. Before you look outside, the chance that your bus is late might be 0.19; once you see that it is raining, the chance is 0.4. The conditional probability P(A | B), read "the probability of A given B", is the probability of A once you know that B has happened. Knowing B throws away every outcome outside B, so B becomes the new sample space, and the probability of A is the share of B that A takes up: P(A ∩ B) ÷ P(B). Two events are independent when knowing one tells you nothing about the other: P(A | B) is the same as P(A), whether or not B happened. That turns into the multiplication rule P(A ∩ B) = P(A)P(B), which is the test you use in an exam.

2The definition: shrink the sample space to B

Here is a gym with its members as the sample space. Let A be the event "a member uses the pool" and B the event "a member attends a group class". Records show P(A) = 0.45, P(B) = 0.30 and P(A ∩ B) = 0.12. The Venn diagram in Figure 1(a) fills in every region: 0.12 in the overlap, 0.45 − 0.12 = 0.33 in A only, 0.30 − 0.12 = 0.18 in B only, and 1 − (0.33 + 0.12 + 0.18) = 0.37 outside both.

Now pick a member who, you are told, attends a class. What is the probability that this member uses the pool? Everyone outside B is now irrelevant, so Figure 1(b) keeps only the circle B, of total probability 0.30. Of that 0.30, the part inside A is 0.12. The answer is the fraction of B that also lies in A:

P(A | B) = P(A ∩ B) ÷ P(B), for P(B) > 0. From the formula booklet.

So 40% of class-goers use the pool, compared with 45% of members overall. Dividing by P(B) rescales B so that it has total probability 1, as U did.

The order matters. P(B | A) is a different question: of the pool users, what fraction attend a class?

Same overlap on top, a different event underneath, a different answer. The event after the bar is always the one you divide by, because it is the one you know has happened. Confusing P(A | B) with P(B | A) is the most expensive mistake in probability; 4.13 (HL) is built on the gap between the two.

Conditional probability from a two-way table. With counts you do not need the formula: restrict attention to the row or column you are given and read the fraction within it. The guide allows this route explicitly. A school of 240 Diploma students records who takes Biology HL and who takes Chemistry HL:

"Given that a student takes Chemistry HL, find the probability that they take Biology HL." Look only at the Chemistry HL column, 60 students, of whom 36 take Biology HL: P(Bio | Chem) = 36/60 = 3/5. The formula gives the same thing, because (36/240) ÷ (60/240) = 36/60. The other way round, P(Chem | Bio) = 36/90 = 2/5.

3The multiplication form, and why trees work

Multiply the definition through by P(B) and you get the form the guide calls the alternate form:

P(A ∩ B) = P(B) P(A | B)

It says: for A and B both to happen, B must happen, and then A must happen given B. That is exactly what you do when you multiply along the branches of a tree diagram, and it is why the second set of branches on any tree carries conditional probabilities.

A commuter's morning bus is late more often in the rain. Let R be "it rains" and L "the bus is late". On a given morning P(R) = 0.3, P(L | R) = 0.4 and P(L | R′) = 0.1. Figure 2 draws the tree. The first branches are P(R) and P(R′); each second branch is a probability given the branch it grows from.

Reading the tree backwards. Now the harder question, which Paper 2 asks often: given that the bus was late, find the probability that it rained. The tree was drawn with rain first, so P(R | L) is not on any branch. Go back to the definition. You know L happened, so L is the new sample space, of probability 0.19; the part of it where it also rained is 0.12.

Knowing the bus was late raises the chance of rain from 0.3 to 0.632, because rain makes lateness four times as likely. The recipe is always the same: the path you want on top, the sum of all the paths consistent with what you know underneath. At HL this recipe, written as a formula, is Bayes' theorem (4.13); at SL you do it exactly as above, straight from the definition.

4Independent events: knowing B tells you nothing about A

Two events are independent when the occurrence of one does not change the probability of the other. In the language of conditional probability:

A and B are independent if P(A | B) = P(A) = P(A | B′).

Read the three parts as a story. Whether B happens (P(A | B)), or B does not happen (P(A | B′)), or you have no idea about B (P(A)), the probability of A is the same number. B carries no information about A.

Figure 3 makes this visible. Each square has area 1, split into a strip for B and a strip for B′, with width P(B) and P(B′). Inside each strip the shaded band is A, and its height is the probability of A within that strip, that is P(A | B) or P(A | B′). In panel (a) the band is level: A takes 30% of B and 30% of B′, so knowing which strip you are in tells you nothing. In panel (b) the band steps down: A takes 60% of B but only 20% of B′, so learning that B happened triples the chance of A.

From the definition to the multiplication rule. If P(A | B) = P(A), put that into the alternate form:

For independent events, P(A ∩ B) = P(A) P(B). From the formula booklet.

In panel (a) of Figure 3 the A-and-B rectangle has width 0.4 and height 0.3, so its area is 0.4 × 0.3 = 0.12 = P(A)P(B). In panel (b) it is 0.4 × 0.6 = 0.24, while P(A) = 0.24 + 0.6 × 0.2 = 0.36 and P(A)P(B) = 0.36 × 0.4 = 0.144. They differ, so those events are not independent.

Why P(A | B′) comes along for free. If P(A ∩ B) = P(A)P(B), then the part of A outside B is P(A ∩ B′) = P(A) − P(A)P(B) = P(A)(1 − P(B)) = P(A)P(B′). Divide by P(B′):

So the three statements P(A | B) = P(A), P(A | B′) = P(A) and P(A ∩ B) = P(A)P(B) stand or fall together, and the same argument shows that if A and B are independent, so are A and B′, A′ and B, and A′ and B′. Independence is a two-way street as well: if P(A | B) = P(A), then P(B | A) = P(A ∩ B) ÷ P(A) = P(A)P(B) ÷ P(A) = P(B).

Where independence comes from. Either the question gives it as a modelling assumption (two separate machines, two dice), and you multiply; or it hands you numbers and asks whether the events are independent, and you test. That is the next section.

5Testing for independence

To test, compute two things that must be equal if the events are independent, and compare them. Any one of these three comparisons is a valid test; use whichever the data make easiest.

Example 1 · from probabilities. For the gym, P(A) = 0.45, P(B) = 0.30 and P(A ∩ B) = 0.12.

Equally, P(A | B) = 0.4 ≠ 0.45 = P(A). A full-marks answer shows the two numbers, the ≠ sign and the conclusion in words; "not independent" alone scores nothing.

Example 2 · from a two-way table, in a medical study. Medical researchers use exactly this test to decide whether something is a risk factor: if a disease is more likely among people with some habit than among people without it, the two are not independent. The data below are invented for this page. A clinic records, for 500 adults, whether each smokes (S) and whether each has a persistent cough (C).

Figure 4 puts the three figures side by side. A smoker in this group is 0.40 ÷ 0.15 ≈ 2.7 times as likely to have a cough as a non-smoker, and in context that is the sentence the question is fishing for: "Smokers in this sample are more likely to have a persistent cough, so the events are not independent; smoking appears to be a risk factor."

Two cautions for the comment. First, the guide says that at SL a data set is treated as the population unless you are told otherwise, so you compare the numbers exactly: 0.40 ≠ 0.15 means not independent. (In real research a small difference in a sample could be chance, and statisticians use formal tests for that; you are not asked to.) Second, "not independent" does not prove that one event causes the other. Say "associated with" or "more likely among", not "causes".

Example 3 · using independence to find an unknown (Paper 1). Events A and B are independent, with P(A) = x, P(B) = 0.4 and P(A ∪ B) = 0.64. Find x.

Check: P(A ∩ B) = 0.16 and 0.4 + 0.4 − 0.16 = 0.64. The move that earns the method mark is replacing P(A ∩ B) with the product before anything else.

6Independent is not the same as mutually exclusive

Students muddle these two. They mean nearly opposite things.

• Mutually exclusive: A and B cannot happen together. P(A ∩ B) = 0. On a Venn diagram the circles do not overlap.
• Independent: A and B can happen together, and the chance of one is untouched by the other. P(A ∩ B) = P(A)P(B). On a Venn diagram the circles overlap by exactly that amount.

Figure 5 draws both with the same P(A) = 0.3 and P(B) = 0.5.

In panel (a), learning that B happened tells you that A certainly did not: P(A | B) = 0. That is the strongest possible influence, so the events are about as far from independent as events can be. Formally, if P(A) > 0 and P(B) > 0 then P(A)P(B) > 0, but P(A ∩ B) = 0, so the multiplication rule fails. Mutually exclusive events with non-zero probabilities are never independent. In panel (b) the overlap is 0.3 × 0.5 = 0.15, and P(A | B) = 0.15 ÷ 0.5 = 0.3, which is P(A) again.