S1 3 Probability | A Level Maths 9709

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🎯What you need to be able to do

Use the addition law, and recognise mutually exclusive events.
Use the multiplication law, and recognise independent events.
Calculate and use conditional probabilities.
Test formally whether two events are independent.
Use tree diagrams, including sampling without replacement.
Use Venn diagrams and two-way tables.
Use the complement to shorten “at least one” calculations.

📚The mathematics

The laws

\( \mathrm{P}(A') = 1 - \mathrm{P}(A) \)

\( \mathrm{P}(A \cup B) = \mathrm{P}(A) + \mathrm{P}(B) - \mathrm{P}(A \cap B) \)

\( \mathrm{P}(A \cap B) = \mathrm{P}(A)\,\mathrm{P}(B \mid A) \)

\( \mathrm{P}(B \mid A) = \dfrac{\mathrm{P}(A \cap B)}{\mathrm{P}(A)} \)

Worth knowing where this sits: the syllabus says explicit use of the addition formula is not required for Paper 5, and most questions can be answered from a tree diagram or a Venn diagram instead. It is included here because it is short, it makes several standard questions immediate, and a Venn diagram is really just this formula drawn. Nothing is lost by reasoning through the diagram instead if you prefer.

Read \( \cup \) as “or” and \( \cap \) as “and”. The subtraction in the addition law exists because anything in both events would otherwise be counted twice.

Mutually exclusive and independent

These are different ideas, and confusing them is the most damaging error in this topic.

Mutually exclusive: \( \mathrm{P}(A \cap B) = 0 \). The events cannot both happen. Then the addition law simplifies to \( \mathrm{P}(A \cup B) = \mathrm{P}(A) + \mathrm{P}(B) \).
Independent: \( \mathrm{P}(A \cap B) = \mathrm{P}(A)\mathrm{P}(B) \). Knowing that one occurred tells you nothing about the other. Equivalently \( \mathrm{P}(B \mid A) = \mathrm{P}(B) \).

Two events with non-zero probability cannot be both. If they are mutually exclusive then \( A \) happening guarantees \( B \) did not, which is the strongest possible dependence. So “mutually exclusive” and “independent” are close to opposites, not synonyms. Adding probabilities for independent events, or multiplying them for mutually exclusive ones, are the two resulting errors.

To test for independence, compute \( \mathrm{P}(A)\mathrm{P}(B) \) and compare it with \( \mathrm{P}(A \cap B) \). Show both numbers and state the conclusion — the comparison is the answer, and a bare “they are independent” earns nothing.

Conditional probability

\[ \mathrm{P}(B \mid A) = \frac{\mathrm{P}(A \cap B)}{\mathrm{P}(A)} \]

Read \( \mathrm{P}(B \mid A) \) as “the probability of \(B\) given that \(A\) has happened”. Conditioning restricts the sample space: you are no longer looking at everything, only at the part where \(A\) is true. That is why \(\mathrm{P}(A)\) sits on the bottom.

Order matters enormously: \( \mathrm{P}(B \mid A) \) and \( \mathrm{P}(A \mid B) \) are generally very different numbers. In a two-way table, the condition tells you which row or column to restrict to, and the denominator is that row or column total rather than the grand total.

Tree diagrams

The workhorse for sequences of events. The rules:

Multiply along a path to get the probability of that whole sequence.
Add between paths that both satisfy the condition.
Each set of branches from a single node must sum to 1 — a free check at every stage.

Watch whether sampling is with or without replacement. Without replacement, the second set of branches has different probabilities and a denominator one smaller, and the events are not independent. Questions signal this with phrases like “two balls are taken without replacement” or “a second card is drawn”.

“At least one”

Almost always faster through the complement:

\[ \mathrm{P}(\text{at least one}) = 1 - \mathrm{P}(\text{none}) \]

Because “none” is a single case while “at least one” is several. This is worth looking for the moment you see the phrase.

✏️Worked example

A bag contains 7 red and 5 blue counters. Two counters are taken at random without replacement. (a) Find the probability that both are red. (b) Find the probability that the counters are of different colours. (c) Given that the second counter is blue, find the probability that the first was red. (d) Events \(A\) and \(B\) satisfy \( \mathrm{P}(A) = 0.4 \), \( \mathrm{P}(B) = 0.35 \) and \( \mathrm{P}(A \cup B) = 0.61 \). Determine whether \(A\) and \(B\) are independent.

(a) There are 12 counters initially, 11 after one is removed:

\[ \mathrm{P}(RR) = \frac{7}{12} \times \frac{6}{11} = \frac{42}{132} = \frac{7}{22} \]

(b) Different colours means red-then-blue or blue-then-red. These are separate paths, so add:

\[ \frac{7}{12}\times\frac{5}{11} + \frac{5}{12}\times\frac{7}{11} = \frac{35}{132} + \frac{35}{132} = \frac{70}{132} = \frac{35}{66} \]

(c) This is conditional, with the condition on the second draw. First find \( \mathrm{P}(\text{2nd is blue}) \), which happens via red-then-blue or blue-then-blue:

\[ \frac{7}{12}\times\frac{5}{11} + \frac{5}{12}\times\frac{4}{11} = \frac{35}{132} + \frac{20}{132} = \frac{55}{132} = \frac{5}{12} \]

Then

\[ \mathrm{P}(\text{1st red} \mid \text{2nd blue}) = \frac{35/132}{55/132} = \frac{35}{55} = \frac{7}{11} \]

(d) From the addition law, \( 0.61 = 0.4 + 0.35 - \mathrm{P}(A \cap B) \), so \( \mathrm{P}(A \cap B) = 0.14 \).

Now compare: \( \mathrm{P}(A)\mathrm{P}(B) = 0.4 \times 0.35 = 0.14 \). Since this equals \( \mathrm{P}(A \cap B) \), the events are independent.

Check it. In (c), notice \( \mathrm{P}(\text{2nd is blue}) = \tfrac{5}{12} \) — exactly the same as the probability the first is blue. That is not a coincidence: without any information about the first draw, every position is equally likely to hold any counter. Spotting it saves the whole calculation. A useful check on (a) and (b): the three outcomes RR, BB and mixed must sum to 1, and \( \tfrac{7}{22} + \tfrac{5}{33} + \tfrac{35}{66} = \tfrac{21 + 10 + 35}{66} = 1 \) ✓.

Conditioning on a later event still works. Part (c) conditions the first draw on the second, which feels backwards but is perfectly legitimate — conditional probability is about information, not chronology. Build the tree forwards as usual, then pick out the paths where the second is blue and take the fraction of those in which the first was red.

📝Practise

Work through these, then reveal the answer.

1. \( \mathrm{P}(A) = 0.5 \), \( \mathrm{P}(B) = 0.3 \), \( \mathrm{P}(A \cap B) = 0.2 \). Find \( \mathrm{P}(A \cup B) \) and \( \mathrm{P}(A \mid B) \).

\( \mathrm{P}(A \cup B) = 0.5 + 0.3 - 0.2 = 0.6 \). \( \mathrm{P}(A \mid B) = \dfrac{0.2}{0.3} = \dfrac{2}{3} = 0.667 \). Note \( \mathrm{P}(A \mid B) = 0.667 \ne \mathrm{P}(A) = 0.5 \), so these events are not independent.

2. A fair die is thrown 4 times. Find the probability of at least one six.

Use the complement. \( \mathrm{P}(\text{no six on one throw}) = \dfrac{5}{6} \), and the throws are independent, so \( \mathrm{P}(\text{no sixes at all}) = \left(\dfrac{5}{6}\right)^{4} = \dfrac{625}{1296} = 0.482 \). Hence \( \mathrm{P}(\text{at least one six}) = 1 - 0.482 = 0.518 \).

3. In a group, 60% study French, 40% study German, and 25% study both. Find the probability that a randomly chosen student studies neither.

\( \mathrm{P}(F \cup G) = 0.6 + 0.4 - 0.25 = 0.75 \). So \( \mathrm{P}(\text{neither}) = 1 - 0.75 = 0.25 \). A Venn diagram makes this immediate: the regions are 0.35 French only, 0.25 both, 0.15 German only, and 0.25 outside, summing to 1 ✓.

4. Two machines produce 60% and 40% of output, with defect rates 3% and 5% respectively. An item is found to be defective. Find the probability it came from the first machine.

\( \mathrm{P}(\text{defective}) = 0.6(0.03) + 0.4(0.05) = 0.018 + 0.020 = 0.038 \). Then \( \mathrm{P}(\text{machine 1} \mid \text{defective}) = \dfrac{0.018}{0.038} = 0.474 \). Although machine 1 makes more items, it is the more reliable one, so it accounts for slightly under half the defects.

5. A box has 4 white and 6 black balls. Three are drawn without replacement. Find the probability that exactly two are white.

The favourable orders are WWB, WBW and BWW, each with the same probability: \( \dfrac{4}{10}\times\dfrac{3}{9}\times\dfrac{6}{8} = \dfrac{72}{720} = 0.1 \). There are 3 such orders, so the probability is \( 3 \times 0.1 = 0.3 \). (Equivalently, by combinations: \( \dfrac{^{4}C_2 \times {}^{6}C_1}{^{10}C_3} = \dfrac{6 \times 6}{120} = 0.3 \) ✓.)

6. \( \mathrm{P}(A) = 0.45 \), \( \mathrm{P}(B) = 0.6 \), \( \mathrm{P}(A \cup B) = 0.78 \). Determine whether \(A\) and \(B\) are independent.

From the addition law, \( 0.78 = 0.45 + 0.6 - \mathrm{P}(A \cap B) \), so \( \mathrm{P}(A \cap B) = 0.27 \). Compare with \( \mathrm{P}(A)\mathrm{P}(B) = 0.45 \times 0.6 = 0.27 \). They are equal, so \(A\) and \(B\) are independent. Showing both numbers and stating the comparison is what earns the marks.

🔗Go deeper — other people’s work

These are external resources, not mine. If one stops working, tell me and everything above it on this page still stands.

Seeing Theory (Brown University) — interactive conditional probability
3Blue1Brown — the Bayes’ theorem video, for why conditioning reverses so counter-intuitively
Khan Academy — tree diagrams and independence