AP Statistics · Unit 2 · Topic 2.6

Conditional Probability Study Lab

A conditional probability is the probability of one event given that another has happened: P(A | B) = P(A and B) ÷ P(B). Set the branch probabilities on a tree, choose a question, and follow the highlighted paths. The table of counts shows the same calculation with whole numbers.

1. Trees, tables, and "given"

Pick a situation, adjust the branch probabilities, and choose what to find.

P(A | B) = P(A and B) ÷ P(B) P(A and B) = P(A) · P(B | A)
Situations

Find

Tree diagram highlighted paths answer the question

Show your work

The same thing with counts out of 10,000

2. Key ideas

Definition

Shrink the sample space

"Given B" means you only look at outcomes where B happened. P(A | B) is the fraction of those outcomes that are also in A: P(A and B) ÷ P(B).

Multiplication rule

Multiply along a branch

Rearranging the definition gives P(A and B) = P(A) · P(B | A). On a tree, multiply the probabilities along a path to get the probability of that whole path.

Tree diagrams

Branches add to 1

The branches leaving any point add to 1. The second-stage branches are conditional probabilities, because they assume the first-stage outcome already happened.

Total probability

Add the paths

To find P(B), add every path that ends in B: P(B) = P(A)·P(B | A) + P(Aᶜ)·P(B | Aᶜ).

Direction matters

P(A | B) ≠ P(B | A)

The probability of a positive test given disease can be 0.95, while the probability of disease given a positive test is far smaller when the disease is rare. Always check which event is “given.”

From tables

Divide by the group

With a two-way table, P(A | B) is the count in A and B divided by the count in B, the same as a conditional relative frequency.

3. Worked examples

4% of a population has a condition. A test is positive for 90% of people with it and 8% of people without it.

FindFormulaWorkAnswer
P(condition and positive)P(C) · P(+ | C)0.04 × 0.900.036
P(no condition and positive)P(Cᶜ) · P(+ | Cᶜ)0.96 × 0.080.0768
P(positive)Add the positive paths0.036 + 0.07680.1128
P(condition | positive)P(C and +) ÷ P(+)0.036 ÷ 0.1128about 0.319
P(no condition | negative)P(Cᶜ and −) ÷ P(−)(0.96 × 0.92) ÷ (1 − 0.1128)about 0.995

4. On the FRQ

How to earn the points

  • Define your events with letters and words: "Let D = has the disease, T = tests positive."
  • Write the formula with the right events, then substitute: "P(D | T) = P(D and T) / P(T) = 0.036 / 0.1128 ≈ 0.319."
  • Draw and label a tree, with probabilities on every branch, when the problem has two stages.
  • Interpret in context: "About 32% of people who test positive actually have the condition."

Common mistakes

  • Swapping the condition: using P(T | D) when the question asks for P(D | T).
  • Forgetting the second path when finding P(T); positives come from both people with and without the condition.
  • Multiplying P(A) · P(B) for "A and B" when the events are not independent. Use P(A) · P(B | A).
  • Dividing by the grand total instead of the “given” group in a two-way table.

5. Practice questions

AP-style multiple choice. Pick an answer to see the explanation.

0 / 0 correct

6. Key terms

The vocabulary for topic 2.6. Read the list, flip through flashcards, or quiz yourself. For every term in the unit, see the Unit 2 terminology guide.