AP® Statistics review sheet from Aim for Five (aimforfive.com/stats/units/2/2-6)
Unit 2 · Topic 2.6
2.6 Conditional Probability
A conditional probability is the chance of one event once you know another has happened. You'll compute it from two-way tables, with the formula P(A | B) = P(A ∩ B)/P(B) and with tree diagrams, and use the general multiplication rule for "and" probabilities in multistage problems.
Key terms
- conditional probability P(A | B)
- general multiplication rule
- tree diagram
- two-way table
What "given" means
P(A | B) is read "the probability of A given B." Knowing B happened shrinks the sample space to just the outcomes in B. Then you ask what share of those are also in A.
From the grade-and-sport table (200 students): P(sport | freshman) = 72/120 = 0.60. You only look at the 120 freshmen, and 72 of them play.
The formula
P(A | B) = P(A ∩ B) ÷ P(B). Using the table: P(sport ∩ freshman) = 72/200 = 0.36 and P(freshman) = 120/200 = 0.60, so P(sport | freshman) = 0.36/0.60 = 0.60. Same answer as restricting to the row.
This formula is on the formula sheet. The condition, the event after the bar, always goes in the denominator.
General multiplication rule
Rearranging gives P(A ∩ B) = P(A) · P(B | A). In words: the chance both happen is the chance the first happens times the chance the second happens given the first.
This is how you handle sequences. Drawing 2 cards without replacement, P(both aces) = (4/52)(3/51) = 12/2652 ≈ 0.0045. After one ace is gone, only 3 aces remain among 51 cards.
Table or tree?
Use a two-way table when you're given counts for two variables. Use a tree when the process happens in stages, or when you're given percentages that are already conditional ("95% of people with the condition test positive").
You can turn a tree into a table by imagining a round number of people. Picture 10,000 people where 2% have a condition: 200 have it and 9,800 don't. If the test catches 95% of real cases, 190 of the 200 test positive. If it wrongly flags 8% of healthy people, 784 of the 9,800 test positive. Then P(condition | positive) = 190/(190 + 784) ≈ 0.195. Thinking in counts makes these problems much easier to reason about.
Tree diagrams
A tree diagram lays out a multistage process. The first set of branches shows the first event's outcomes with their probabilities. From each branch, the second set shows conditional probabilities given that first outcome.
Multiply along a path to get the probability of that path (an "and" probability). Add the paths that make up an event to get its total probability. The branches from any one point add to 1.
Check a finished tree two ways: the branches leaving each point should add to 1, and the probabilities of all the complete paths should also add to 1.
To reverse a condition, for example from P(positive | disease) to P(disease | positive), use the tree: P(disease | positive) = P(disease ∩ positive) ÷ P(positive), where P(positive) adds every path ending in a positive result.
Worked examples
Try each one yourself first, then open the solution.
- Example 1Calculator allowed
Medical test with a tree diagram
2% of a population has a certain condition. A screening test is positive for 95% of people with the condition and for 8% of people without it. A randomly chosen person tests positive. What's the probability they have the condition?
Show the solutionHide the solution
- Step 1: First branches: condition (0.02) and no condition (0.98).
- Step 2: Second branches: given condition, positive 0.95; given no condition, positive 0.08.
- Step 3: P(condition ∩ positive) = 0.02 × 0.95 = 0.019.
- Step 4: P(no condition ∩ positive) = 0.98 × 0.08 = 0.0784.
- Step 5: P(positive) = 0.019 + 0.0784 = 0.0974.
- Step 6: P(condition | positive) = 0.019 ÷ 0.0974 ≈ 0.195.
Answer: About 0.195. Only about 19.5% of people who test positive actually have the condition, because the condition is rare and false positives from the large healthy group outnumber the true positives.
- Example 2Calculator allowed
Trap: reversing the condition
From the grade-and-sport table, a student is asked for the probability that a randomly chosen sport player is a freshman and answers 72/120 = 0.60. Find the correct answer.
Show the solutionHide the solution
- Step 1: "A sport player is a freshman" means P(freshman | sport). The condition is "plays a sport."
- Step 2: Restrict to the 100 sport players. 72 of them are freshmen.
- Step 3: P(freshman | sport) = 72/100 = 0.72. The student computed P(sport | freshman) instead.
Answer: 0.72. The student swapped the condition and the event.
Common mistakes
- Mixing up P(A | B) and P(B | A). They're usually different.
- Using the grand total as the denominator instead of the total for the given condition.
- Forgetting to update probabilities on the second draw when sampling without replacement.
- Adding along a tree path instead of multiplying.
On the exam
- One multiple-choice set on the exam focuses on probability and random variables, so expect several connected questions built on one table or tree.
- On free response, show the formula with numbers substituted, such as P(A | B) = 0.019/0.0974, so a reader can follow your work.
Connected topics
Videos
Check yourself
4 questions on 2.6 Conditional Probability. Pick an answer to see if you got it, and why.
A screening test is used for a condition that 2% of a population has. If a person has the condition, the test is positive 95% of the time. If a person doesn't have the condition, the test is still positive 8% of the time.
Described scenario with invented probabilities
What is the probability that a randomly selected person from this population tests positive?
Given that a randomly selected person tests positive, what is the probability that the person actually has the condition?
What is the probability that a randomly selected person has the condition and tests negative?
A club has 5 seniors and 7 juniors. Two members are chosen at random, without replacement, to attend a conference. What is the probability that both are seniors?
0 of 4 answered