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.
Today's hook, A 99% accurate test says you have a rare disease. Should you panic? The actual chance you have the disease could be just 50%. Conditional probability explains this shocking result, and why it has sent innocent people to prison.
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Worksheets
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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.
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The two moves
Work through the core explanation before applying it.
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The two moves
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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.
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 $P(\text{tennis} \mid \text{basketball})$?
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.
The prosecutor's fallacy. $P(A \mid B)$ and $P(B \mid A)$ are generally very different. $P(\text{disease} \mid \text{positive}) = 0.5$ but $P(\text{positive} \mid \text{disease}) = 0.99$. Confusing them has caused wrongful convictions. Always write the condition on the right side of $\mid$ clearly.
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
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Work examples end to end
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PROBLEM 1 · CONDITIONAL FORMULA
If $P(A \cap B) = 0.2$ and $P(B) = 0.5$, find $P(A \mid B)$.
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$P(A \mid B) = \dfrac{P(A \cap B)}{P(B)}$
Write the conditional probability formula.
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$= \dfrac{0.2}{0.5} = 0.4$
Substitute and simplify. $P(A \mid B) = 0.4$.
PROBLEM 2 · TOTAL PROBABILITY
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 $P(\text{from X} \mid \text{defective})$.
Use the conditional formula. 75% of defective items came from Supplier X despite X only supplying 60% of components, because X has double the defect rate.
PROBLEM 3 · TWO-WAY TABLE
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 $P(\text{sport} \mid \text{work})$.
Now restrict to the "work" row (80 students). Of these, 45 play sport. Denominator is the row total.
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
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Dodge the traps, then apply
Meet the mistakes that cost marks, then do it yourself.
Trap 01
Swapping $P(A \mid B)$ with $P(B \mid A)$
$P(\text{disease} \mid \text{positive}) = 0.5$ but $P(\text{positive} \mid \text{disease}) = 0.99$. These are completely different values. Always write the condition clearly on the right side of $\mid$.
Trap 02
Using the grand total as denominator
In a two-way table, $P(\text{work} \mid \text{sport})$ uses the "sport" total (115), not the grand total (200). Conditioning means you have restricted your world to just those outcomes.
Trap 03
Adding instead of applying total probability
$P(\text{choir}) \neq P(\text{choir} \mid \text{player}) + P(\text{choir} \mid \text{non-player})$. You must weight by $P(\text{player})$ and $P(\text{non-player})$ first, then add.
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:
$P(A \mid B)$
$P(A \cap B)$
$P(A \cup B)$
Prob of A and B both occurring
Prob of A or B occurring
Prob of A given B has occurred
Drill activities
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Run the quick drill and copy the summary into your book.
A bag has 5 red and 5 blue marbles. Two drawn without replacement. Given first is red, find $P(\text{second red})$.
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100 people: 30 have a passport, 20 have both passport and licence, 60 have a licence. Find $P(\text{passport} \mid \text{licence})$.
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Factory: three machines produce 50%, 30%, 20% of output with defect rates 2%, 3%, 5%. Find $P(\text{defective})$.
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Drug test: 95% accurate, 5% of athletes use drug. Find $P(\text{user} \mid \text{positive})$.
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Revisit your thinking
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 $A$.
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Multiple choice
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Multiple choice
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Short answer
ApplyBand 43 marks
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)
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ApplyBand 43 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)
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AnalyseBand 53 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)
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].
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quiz
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