Knowing that one event has happened shrinks the sample space. Conditional probability measures the chance of another event inside that smaller space, and comparing the two tells you whether the events are independent.
\(\text{P}(A \mid B) = \dfrac{\text{P}(A \cap B)}{\text{P}(B)}\): the probability of \(A\) once you know \(B\) has happened. Events are independent exactly when \(\text{P}(A \cap B) = \text{P}(A) \times \text{P}(B)\), which is the same as \(\text{P}(A \mid B) = \text{P}(A)\).
Fig. 1: the tree is drawn forwards (weather, then lateness). Given that the bus is late, only the two 'Late' routes remain, and the rain route is 0.12 of their 0.19
Section 01
Given that.
In the lab's conditional tab, click on a condition and watch the sample space shrink to just those outcomes. The new probability is the share of that smaller space taken by the event you want. In a two-way table, 'given' simply tells you which row or column to use as the total.
French
Not French
Total
Male
18
42
60
Female
30
30
60
Total
48
72
120
Worked example From a two-way table
The table shows 120 students and whether they study French. One student is chosen at random. Find (a) P(studies French | female), (b) P(female | studies French).
(a) 'Given female' restricts us to the 60 females, of whom 30 study French: \(\dfrac{30}{60}\).
(b) 'Given French' restricts us to the 48 French students, of whom 30 are female: \(\dfrac{30}{48}\).
With the formula: \(\text{P}(F \mid \text{French}) = \dfrac{30/120}{48/120} = \dfrac{30}{48}\). Same answer.
Answer: (a) \(\dfrac{1}{2}\) (b) \(\dfrac{5}{8}\)
Examiner trap
Mixing up \(\text{P}(A \mid B)\) and \(\text{P}(B \mid A)\). They have the same numerator but different denominators: \(\tfrac{1}{2}\) and \(\tfrac{5}{8}\) above.
Section 02
Reversing a tree.
Trees are drawn in time order, so their second branches are conditional probabilities such as P(late | rain). When a question asks the reverse, such as P(rain | late), use the formula: the route you want divided by all the routes that satisfy the condition.
Worked example Was it raining?
The probability that it rains on a given day is 0.3. If it rains, the probability that the bus is late is 0.4; if it is dry, it is 0.1. Given that the bus is late, find the probability that it rained.
Draw the tree (Fig. 1). Rain and late: \(0.3 \times 0.4 = 0.12\). Dry and late: \(0.7 \times 0.1 = 0.07\).
2% of people have a condition. A test is positive for 95% of people with the condition and for 8% of people without it. Find the probability that a person who tests positive has the condition.
Condition and positive: \(0.02 \times 0.95 = 0.019\).
No condition and positive: \(0.98 \times 0.08 = 0.0784\).
\(\text{P}(\text{positive}) = 0.0974\), so \(\text{P}(\text{condition} \mid \text{positive}) = \dfrac{0.019}{0.0974}\).
Answer: 0.195 (3 s.f.): most positive results are false alarms
Your turn
A bag has 5 red and 3 blue balls. Two are taken without replacement. Given that the second ball is red, find the probability that the first was red.
Answering 0.4. That is P(late | rain), read straight off the branch; the question asks for P(rain | late), which needs the whole calculation.
Stage 1 · Learn
Check your understanding
Four quick checks. Each correct answer earns XP and records this skill on your profile.
Quick check+10 XP
\(\text{P}(A \cap B) = 0.12\) and \(\text{P}(B) = 0.4\). Find \(\text{P}(A \mid B)\).
Quick check+10 XP
If \(A\) and \(B\) are independent, which statement is true?
Quick check+10 XP
\(\text{P}(A) = 0.5\), \(\text{P}(B) = 0.4\) and \(\text{P}(A \cap B) = 0.2\). Are \(A\) and \(B\) independent?
Quick check+10 XP
\(A\) and \(B\) are exclusive, with \(\text{P}(A) = 0.2\) and \(\text{P}(B) = 0.5\). Are they independent?
Section 03
Testing independence.
To decide whether two events are independent, calculate \(\text{P}(A \cap B)\) and \(\text{P}(A) \times \text{P}(B)\) separately and compare. Equal means independent; different means not. Show both numbers: the comparison IS the answer. The Venn tab of the lab shows the same check with areas.
Worked example Two dice
Two fair dice are thrown. \(A\): the first die is even. \(B\): the total is 7. \(C\): the total is 8. Is \(A\) independent of \(B\)? Of \(C\)?
\(A \cap B\): (2,5), (4,3), (6,1), so \(\tfrac{3}{36} = \tfrac{1}{12}\). And \(\tfrac{1}{2} \times \tfrac{1}{6} = \tfrac{1}{12}\). Equal.
\(A \cap C\): (2,6), (4,4), (6,2), so \(\tfrac{3}{36} = \tfrac{1}{12}\). But \(\tfrac{1}{2} \times \tfrac{5}{36} = \tfrac{5}{72}\). Not equal.
Answer: \(A\) and \(B\) are independent; \(A\) and \(C\) are not
Worked example From a Venn diagram
\(\text{P}(A) = 0.4\), \(\text{P}(B) = 0.5\) and \(\text{P}(A \cup B) = 0.7\). Show that \(A\) and \(B\) are independent and find \(\text{P}(A \mid B)\).
In a Venn diagram, \(\text{P}(A) + \text{P}(B)\) counts the overlap twice: \(0.4 + 0.5 - 0.7 = 0.2 = \text{P}(A \cap B)\).
\(\text{P}(A \mid B) = \dfrac{0.2}{0.5} = 0.4\), which equals \(\text{P}(A)\), as independence predicts.
Answer: Independent; \(\text{P}(A \mid B) = 0.4\)
Examiner trap
Deciding independence by intuition ('the dice are separate, so the events must be independent'). Events defined on the same throw can go either way: \(A\) is independent of \(B\) but not of \(C\).
Section 04
Exclusive is not independent.
If \(A\) and \(B\) are exclusive, knowing \(B\) happened tells you for certain that \(A\) did not, which is the opposite of independence. So two exclusive events with non-zero probabilities are never independent.
Worked example Both ways
\(\text{P}(A) = 0.3\) and \(\text{P}(B) = 0.4\). Find \(\text{P}(A \cup B)\) when (a) \(A\) and \(B\) are exclusive, (b) \(A\) and \(B\) are independent. Explain why they cannot be both.