1The idea in one paragraph

Two events can overlap, and every rule on this page is about handling the overlap. "A or B" means A, or B, or both, so adding P(A) and P(B) counts the overlap twice and you subtract it once. If the events cannot happen together, they are mutually exclusive, the overlap is empty, and you just add. "A given B" means you already know B happened, so B becomes the whole sample space and you ask what fraction of it is also A. If knowing B makes no difference to the chance of A, the events are independent, and the probability of both is simply the product. Venn diagrams, tables, grids and trees are four ways of seeing the same overlap; draw one and most questions answer themselves.

2Venn diagrams and the language of events

A Venn diagram draws the sample space U as a rectangle and each event as a region inside it. Three symbols name the regions you need, and Figure 1 shades the four you will meet most.

• A ∩ B, the intersection, is "A and B": outcomes in both.
• A ∪ B, the union, is "A or B": outcomes in A, in B, or in both.
• A′, the complement, is "not A": everything in U outside A.
• Combine them freely. A ∩ B′ is "A and not B", the part of A outside B.

Filling in a Venn diagram from counts. Always start in the middle. Sixty students were asked about their subjects: 34 take biology (B), 25 take chemistry (C), and 12 take both. Figure 2 shows the order of work.

The classic mistake is to write 34 in the biology-only region. The 34 is the whole circle, overlap included. Once the diagram is filled, every probability is a count over 60: P(B ∩ C) = 12/60 = 1/5, P(B ∪ C) = 47/60, P(neither) = 13/60, P(exactly one subject) = (22 + 13)/60 = 35/60 = 7/12.

The same diagram works with probabilities in place of counts. The four regions then add to 1 instead of to 60.

3Combined events: the addition rule, and what "or" means

In mathematics, "or" includes both. "The student takes biology or chemistry" is true of a student who takes both. That is the non-exclusivity of "or", and the guide names it because everyday English often means "one or the other but not both". In an exam, "A or B" always means A ∪ B, overlap included. If a question wants one but not both, it will say "exactly one" or "either … but not both".

Adding P(B) and P(C) counts every student in the overlap twice, once in each circle. Subtract the overlap once to fix it.

P(A ∪ B) = P(A) + P(B) − P(A ∩ B)

The rule has four quantities, so any three give the fourth. A typical Paper 1 opening: P(A) = 0.5, P(B) = 0.4 and P(A ∪ B) = 0.7. Find P(A ∩ B).

Then fill a Venn diagram with probabilities, starting in the middle: A only = 0.5 − 0.2 = 0.3, B only = 0.4 − 0.2 = 0.2, neither = 1 − 0.7 = 0.3. The guide is explicit that you may solve problems from a diagram without writing formulae at all, and on hard questions the diagram is often the safer route. We come back to this pair of events in section 8.

4Mutually exclusive events

Two events are mutually exclusive when they cannot happen at the same time: there is no outcome in both. The overlap is empty.

Mutually exclusive: P(A ∩ B) = 0, so P(A ∪ B) = P(A) + P(B).

Figure 3 shows one roll of a die twice. In panel (a), A = {1, 2} and B = {5, 6}: no score is in both, so the circles do not touch and P(A ∪ B) = 2/6 + 2/6 = 4/6. In panel (b), A = "even" = {2, 4, 6} and B = "at least 4" = {4, 5, 6}: they share 4 and 6, so they are not mutually exclusive, and the addition needs the correction term: 3/6 + 3/6 − 2/6 = 4/6. Check by listing: A ∪ B = {2, 4, 5, 6}, four scores.

An event and its complement are always mutually exclusive, and together they fill U. That is why P(A) + P(A′) = 1 in 4.5 is a special case of this rule.

To show that two events are not mutually exclusive, find P(A ∩ B) and show it is not 0, or name one outcome in both. "They can both happen" without evidence earns nothing.

5Tables of outcomes and sample space diagrams

A table of outcomes, also called a two-way table, counts a group sorted two ways at once. It holds exactly the same information as a Venn diagram, arranged in a grid. Here are 120 people at a cinema (invented data), sorted by age and by whether they bought popcorn.

Read "and" from a single cell and "or" by combining cells.

The subtraction of 28 is the addition rule again: the 28 adults with popcorn sit in both the adult row and the popcorn column.

A sample space diagram is the grid from 4.5, one trial across the top and one down the side. It is the right tool when two equally likely trials combine, and you count cells. Section 6 uses one for a conditional probability.

6Conditional probability

A conditional probability is the probability of one event when you already know that another has happened. P(A | B) is read "the probability of A given B".

Knowing B has happened throws away every outcome outside B. What remains, B itself, is the new sample space, and you ask what fraction of it is also in A. Figure 4 shows this with the subject survey: given that a student takes chemistry, only the 25 chemistry students are left, and 12 of them take biology.

Dividing top and bottom by 60 turns counts into probabilities, and gives the rule in the formula booklet.

P(A | B) = P(A ∩ B) ÷ P(B), and rearranged, P(A ∩ B) = P(B) × P(A | B).

Order matters. P(C | B), the chance a biology student takes chemistry, is 12/34 = 6/17, not 12/25. The event after the bar is the one you know, and it goes on the bottom.

From the cinema table: given that a person is a child, what is the probability they bought popcorn? Restrict to the child row, 48 people, of whom 30 bought popcorn: P(popcorn | child) = 30/48 = 5/8. Reversed, given popcorn, the probability of a child is 30/58 = 15/29, restricting to the popcorn column. Same 30 on top, different bottoms.

From a sample space diagram. Two fair dice are rolled. Given that at least one shows a 5, find the probability that the total is 8. Figure 5 marks the 11 cells with a 5 in amber; that is the new sample space. Only two of them, (3, 5) and (5, 3), total 8.

The condition changed the probability, which is the whole point of the idea: new information changes how likely things are.

7Tree diagrams, with and without replacement

A tree diagram shows a sequence of trials. Each trial is a fan of branches, each branch carries its probability, and the branches from any one point add to 1. Two rules drive it.

Multiply along a path to get the probability of that whole sequence. Add the paths that give the event you want.

Multiplying along a path is P(A ∩ B) = P(A) × P(B | A) in picture form: the second branch always carries the probability of the second event given what happened on the first.

With replacement. A bag holds 5 red and 3 green counters. One is taken, its colour noted, and it is put back; then a second is taken. Because the bag is restored, the second draw has exactly the same branches as the first, as Figure 6 shows.

Without replacement. Now the first counter is kept. After a red is taken, the bag holds 4 red and 3 green, 7 in all; after a green, 5 red and 2 green. The second-draw branches change depending on the first, as in Figure 7.

Removing a counter changes both the numerator and the denominator of the next fraction. Forgetting to reduce the 8 to 7 is the commonest single slip in the topic.

A conditional probability read backwards. Trees are drawn in time order, but questions often ask about the first stage given the second. Given that the second counter is green, find the probability that the first was red. Use the definition, with both parts read off the tree.

The same method works in context. On any school day the probability of rain is 0.3. When it rains, Leo is late with probability 0.4; when it does not, with probability 0.1.

Knowing that Leo was late more than doubles the chance that it rained, from 0.3 to 0.632. Finding "the first stage given the second" is exactly what HL students later call Bayes' theorem; at SL you do it with the definition and a tree, as here.

8Independent events

Two events are independent when knowing that one has happened does not change the probability of the other: P(A | B) = P(A). Put that into P(A ∩ B) = P(B) × P(A | B) and the rule appears.

Independent: P(A ∩ B) = P(A) × P(B). This is also the test: if the product equals the probability of both, the events are independent; if not, they are not.

Separate physical trials are the natural case. A coin and a die do not influence each other, so P(head and six) = 1/2 × 1/6 = 1/12. Draws with replacement are independent; draws without replacement are not, because the first draw changes the bag.

Testing for independence. Return to section 3's pair: P(A) = 0.5, P(B) = 0.4, P(A ∩ B) = 0.2.

Now the subject survey: P(B) × P(C) = (34/60) × (25/60) = 17/72 ≈ 0.236, but P(B ∩ C) = 12/60 = 0.2. They differ, so taking biology and taking chemistry are not independent. In the cinema, P(child) × P(popcorn) = 0.4 × 58/120 ≈ 0.193, while P(child and popcorn) = 0.25: not independent either; children are more likely to buy popcorn. When a question says "determine whether", give the two numbers, compare them, and state the conclusion. Page 4.11 returns to independence formally, including the equivalent test P(A | B) = P(A) = P(A | B′).

Using independence: "at least one". A sensor detects a fault with probability 0.7, and three sensors work independently. The probability that at least one detects the fault is best found through the complement: all three miss it with probability 0.3 × 0.3 × 0.3 = 0.027, so P(at least one detects) = 1 − 0.027 = 0.973. This combines 4.5's complement with this section's product rule.

Independent is not the same as mutually exclusive. They sound similar and are almost opposites. Mutually exclusive events cannot happen together, so if one happens the other certainly does not: knowing one changes everything about the other. Two events with non-zero probabilities cannot be both. Check with numbers: if P(A) = 0.3, P(B) = 0.5 and they are mutually exclusive, then P(A ∩ B) = 0, but P(A) × P(B) = 0.15, so they are not independent.