Conditional Probability
A medical test comes back positive. What is the actual probability you have the disease? With a 99% accurate test for a rare condition, a positive result is only 50% reliable. The key is $P(A \mid B)$ — the probability of an event given another has occurred. By the end of this lesson you will solve these counterintuitive problems with the formula, tree diagrams, and two-way tables.
Practise this lesson
Three printable worksheets that build from foundations to mastery — or build your own from any module’s questions.
If $P(A \mid B) = P(A)$, what does this tell us about the relationship between events $A$ and $B$? Make a prediction before reading on — no formula needed yet.
Every conditional probability question this year lives inside one formula. Lock it in now — the rest of this lesson is learning to recognise when and how to use it.
Conditional probability restricts the sample space. Dividing by $P(B)$ re-normalises so probabilities still sum to 1 within the smaller world where $B$ has occurred. The multiplication rule rearranges the same equation.
Key facts
- $P(A \mid B) = \dfrac{P(A \cap B)}{P(B)}$
- $P(A \cap B) = P(B) \times P(A \mid B)$
- Independence means $P(A \mid B) = P(A)$
Concepts
- Conditional probability restricts the sample space to event $B$
- Tree diagrams encode conditional probabilities on branches
- Two-way tables organise joint and marginal probabilities
Skills
- Calculate conditional probabilities from two-way tables
- Build and use tree diagrams for multi-stage experiments
- Apply the total probability formula
The conditional probability of $A$ given $B$, written $P(A \mid B)$, is the probability that $A$ occurs assuming $B$ has already occurred.
Why divide by $P(B)$? Because we restrict our attention to only those outcomes where $B$ happened. Within this smaller universe, the fraction that also satisfies $A$ is $P(A \cap B) / P(B)$.
Example. In a class, 30% of students play basketball and 20% play both basketball and tennis. If you select a basketball player at random, what is $P(\text{tennis} \mid \text{basketball})$?
$$P(\text{tennis} \mid \text{basketball}) = \frac{P(\text{both})}{P(\text{basketball})} = \frac{0.20}{0.30} = \frac{2}{3}$$
Two-thirds of basketball players also play tennis. Notice how the conditional probability ($\frac{2}{3}$) is very different from the unconditional probability of playing tennis (which would be lower).
Multiply along branches to find joint probabilities. $P(B \mid A)$ is the probability of $B$ given $A$ has occurred.
Conditional probability: $P(A \mid B) = \dfrac{P(A \cap B)}{P(B)}$ — dividing by $P(B)$ restricts the sample space to $B$; Multiplication rule: $P(A \cap B) = P(B) \times P(A \mid B)$ — multiply along tree branches
Pause — copy the conditional probability formula $P(A \mid B) = \frac{P(A \cap B)}{P(B)}$ (dividing by $P(B)$ restricts the sample space) and the tree-branch rule: multiply probabilities along branches into your book.
Did you get this? True or false: $P(A \mid B) = \dfrac{P(A \cap B)}{P(B)}$, where $P(B) > 0$.
Worked examples · 3 in a row, reveal as you go
If $P(A \cap B) = 0.2$ and $P(B) = 0.5$, find $P(A \mid B)$.
A company sources components from two suppliers: 60% from Supplier X (4% defective) and 40% from Supplier Y (2% defective). (a) Find $P(\text{defective})$. (b) Find $P(\text{from X} \mid \text{defective})$.
200 students were surveyed on part-time work and sport. From the table: 45 work and play sport, 80 work total, 115 play sport total. Find $P(\text{work} \mid \text{sport})$ and $P(\text{sport} \mid \text{work})$.
Quick check: A school has 40% instrument players. Of these, 25% sing in choir. Of non-players, 10% sing. What is $P(\text{choir})$?
Common errors · the traps that cost marks
Fill in the blank: In the law of total probability, $P(A) = P(B) \times P(A \mid B) + P(B') \times P(A \mid \underline{\hspace{60px}})$.
Match each expression to its meaning:
Drill activities
$P(A \cap B) = 0.12$, $P(B) = 0.4$. Find $P(A \mid B)$.
A bag has 5 red and 5 blue marbles. Two drawn without replacement. Given first is red, find $P(\text{second red})$.
100 people: 30 have a passport, 20 have both passport and licence, 60 have a licence. Find $P(\text{passport} \mid \text{licence})$.
Factory: three machines produce 50%, 30%, 20% of output with defect rates 2%, 3%, 5%. Find $P(\text{defective})$.
Drug test: 95% accurate, 5% of athletes use drug. Find $P(\text{user} \mid \text{positive})$.
Earlier you were asked: if $P(A \mid B) = P(A)$, what does this tell us? It means knowing $B$ occurred does not change the probability of $A$. This is the definition of independence: $A$ and $B$ are independent if and only if $P(A \mid B) = P(A)$ — equivalently, $P(A \cap B) = P(A) \times P(B)$. $B$ gives us no information about $A$.
Pick your answer, then rate your confidence — that tells the system what to drill next. Each retry pulls a fresh mix from the bank.
Q1. In a school, 40% of students play a musical instrument. Of those who play an instrument, 25% also sing in the choir. Of those who do not play an instrument, 10% sing in the choir. (a) Draw a tree diagram. (b) Find the probability that a randomly selected student sings in the choir. (c) Given that a student sings in the choir, find the probability they play an instrument. (3 marks)
Q2. A survey of 500 people on diet and exercise:
| Regular Exercise | Irregular Exercise | Total | |
|---|---|---|---|
| Healthy Diet | 120 | 80 | 200 |
| Unhealthy Diet | 90 | 210 | 300 |
| Total | 210 | 290 | 500 |
(a) Find $P(\text{healthy diet} \mid \text{regular exercise})$. (b) Find $P(\text{regular exercise} \mid \text{healthy diet})$. (c) Determine whether diet and exercise are independent, justifying with calculations. (3 marks)
Q3. A disease affects 2% of a population. A test has a 98% true positive rate and a 3% false positive rate. (a) In a population of 10 000, construct a two-way table showing expected numbers of true positives, false positives, true negatives, and false negatives. (b) Calculate $P(\text{disease} \mid \text{positive})$. (c) A politician claims: "This test is 98% accurate, so if you test positive, you almost certainly have the disease." Analyse this claim mathematically and explain why it is misleading. (3 marks)
Comprehensive answers (click to reveal)
Drill 1: $0.3$ · 2: $\frac{4}{9}$ · 3: $\frac{1}{3}$ · 4: $0.029$ · 5: $0.5$
Q1 (3 marks): (a) Tree: Instrument (0.4) → Choir (0.25), No choir (0.75); No instrument (0.6) → Choir (0.10), No choir (0.90) [0.5]. (b) $P(\text{choir}) = 0.4 \times 0.25 + 0.6 \times 0.10 = 0.10 + 0.06 = 0.16$ [1]. (c) $P(\text{instrument} \mid \text{choir}) = \frac{0.10}{0.16} = \frac{5}{8} = 0.625$ [1.5].
Q2 (3 marks): (a) $P(\text{healthy} \mid \text{regular}) = \frac{120}{210} = \frac{4}{7} \approx 0.571$ [0.5]. (b) $P(\text{regular} \mid \text{healthy}) = \frac{120}{200} = 0.6$ [0.5]. (c) $P(\text{healthy}) \times P(\text{regular}) = 0.4 \times 0.42 = 0.168$. But $P(\text{healthy} \cap \text{regular}) = \frac{120}{500} = 0.24$. Since $0.24 \neq 0.168$, not independent [2].
Q3 (3 marks): (a) Disease+: 200 people; TP = 196, FN = 4. Disease−: 9 800; FP = 294, TN = 9 506 [0.5]. (b) $P(\text{disease} \mid \text{positive}) = \frac{196}{490} = 0.4 = 40\%$ [1]. (c) The claim confuses $P(\text{positive} \mid \text{disease}) = 0.98$ with $P(\text{disease} \mid \text{positive}) = 0.40$. Prevalence matters: in a low-prevalence population, false positives from the healthy majority outnumber true positives from the sick minority. The public needs to know the base rate (prevalence) to correctly interpret any test result [1.5].
Five timed questions. Beat the boss to bank a tier — gold (90% + speed), silver (75%), or bronze (50%). Replays welcome.
⚔ Enter the arenaClimb platforms by answering conditional probability questions. Lighter alternative to the boss.
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