Conditional Probability — AP® Statistics Lesson

Part 1 · Hook

Why this matters

A screening test catches 90% of people who have a condition. You test positive. Is there a 90% chance you have it? Surprisingly, no: in a typical population it is closer to 38%. The difference is which group a probability is about, and a tree diagram makes it visible.

Part 2 · Before you start

What this builds on

Part 3 · Prerequisite check

Quick check before you start

1. Of 150 returning park visitors, 54 fished. What is the conditional relative frequency of fishing for returning visitors?

0.36
0.216
0.783
0.54

Show the answer

$\frac{54}{150} = 0.36$ : divide by the group's total.

Correct: 0.36:
0.216:
0.783:
0.54:

2. What is $P(A \cap B)$ ?

The probability that A and B both happen
The probability that A or B happens
The probability of A minus B
The probability that neither happens

Show the answer

$\cap$ is "and": both events.

Correct: The probability that A and B both happen:
The probability that A or B happens:
The probability of A minus B:
The probability that neither happens:

Part 4 · See it

See it first

Tree diagram for a screening test. First branches: 0.04 of people have the condition and 0.96 do not. Given the condition, the test is positive with probability 0.90 and negative 0.10. Given no condition, positive 0.06 and negative 0.94. The four paths multiply to 0.036, 0.004, 0.0576 and 0.9024. The positive paths add to 0.0936. — First branch: condition or not. Second branch: the test result given the first. Multiply along a path; add the paths you want. LevlPrep original diagram.

Part 5 · Step by step

How it works, step by step

You are told that B happenedonly the outcomes in B are possible now, so B becomes the new total
So the share of B that is also Ais the conditional probability: $P(A \mid B) = \frac{P(A \cap B)}{P(B)}$
Rearrange the rulethe multiplication rule: $P(A \cap B) = P(A) \cdot P(B \mid A)$ , which is multiplying along a tree's path
Add every path that ends in the result you wantyou get its total probability, and dividing one path by it reverses the condition

Part 6 · Key ideas

Key ideas

A conditional probability $P(A \mid B)$ is the probability of A given that B happened: $P(A \mid B) = \frac{P(A \cap B)}{P(B)}$ . In a table, divide by the total of B's row or column.
The multiplication rule: $P(A \cap B) = P(A) \cdot P(B \mid A)$ . Drawing without replacement changes the second probability.
A tree diagram shows stages: first-stage probabilities, then conditional ones. Multiply along a path; add paths.
$P(A \mid B)$ and $P(B \mid A)$ are usually different. Name the group after the bar.

Part 7 · Misconception

A common mistake

The wrong idea: If the test is positive for 90% of people with the condition, then 90% of people who test positive have it.

What actually happens: Those are different conditional probabilities: $P(\text{positive} \mid \text{condition})$ $= 0.90$ , but $P(\text{condition} \mid \text{positive})$ $= \frac{0.036}{0.0936}$ $\approx 0.385$ .

Part 8 · Check yourself

Check yourself

Exam-style questions. Anything you miss goes into your review queue.

Table

Main news source by age group

A random sample of 400 adults in a city were asked their main source of news. One of these adults is chosen at random (invented data).

Age group and main news source for 400 adults
Age group	Online	TV	Print	Total
Under 40	132	66	22	220
40 and over	54	90	36	180
Total	186	156	58	400

1. What is the probability that the adult gets news mainly online, given that the adult is under 40?

0.33
0.710
0.6
0.465

Show the answer

Given under 40, the 220 adults under 40 are the new total: $P(\text{online} \mid \text{under } 40)$ $= \frac{132}{220}$ $= 0.6$ .

0.33: That is $\frac{132}{400}$ , the joint probability of under 40 and online.
0.710: That is $\frac{132}{186}$ , the probability of under 40 given online: the condition is reversed.
Correct: 0.6: Right: 132 of the 220 adults under 40.
0.465: That is $\frac{186}{400}$ , the probability of online for everyone.

2. Given that the adult gets news mainly from TV, what is the probability that the adult is under 40?

0.3
0.165
0.55
0.423

Show the answer

The condition is TV, so divide by the 156 TV viewers: $P(\text{under } 40 \mid \text{TV})$ $= \frac{66}{156}$ $\approx 0.423$ .

0.3: That is $\frac{66}{220}$ , the probability of TV given under 40: the condition is reversed.
0.165: That is $\frac{66}{400}$ , a joint probability.
0.55: That is $\frac{220}{400}$ , the probability of under 40 for everyone.
Correct: 0.423: Right: 66 of the 156 TV viewers.

3. What is the probability that the adult is 40 or over and gets news mainly from print?

0.09
0.2
0.621
0.145

Show the answer

This is a joint probability, one cell over all 400: $P(40 \text{ or over} \cap \text{print})$ $= \frac{36}{400}$ $= 0.09$ .

Correct: 0.09: Right: 36 of all 400 adults.
0.2: That is $\frac{36}{180}$ , a conditional probability given 40 or over.
0.621: That is $\frac{36}{58}$ , a conditional probability given print.
0.145: That is $\frac{58}{400}$ , the probability of print for everyone.

Diagram

A screening test

A screening test is used on a large population in which 4% of people have a certain condition. For people with the condition, the test is positive 90% of the time. For people without it, the test is positive 6% of the time. One person from the population is tested.

4. What is the probability that the test is positive?

Type a number. A table answer and a calculator answer are both accepted.

Show the answer

Add the two branches that end in positive: $P(\text{positive})$ $= (0.04)$ $(0.90)$ $+ (0.96)$ $(0.06)$ $= 0.036$ $+ 0.0576$ $= 0.0936$ .

Answer: 0.0936

5. Given that the test is positive, what is the probability that the person has the condition?

Type a number. A table answer and a calculator answer are both accepted.

Show the answer

$P(\text{condition} \mid \text{positive})$ $= \frac{P(\text{condition} \cap \text{positive})}{P(\text{positive})}$ $= \frac{0.036}{0.0936}$ $\approx 0.385$ .

Answer: 0.3846

6. A bag has 5 red and 3 blue tokens. Two tokens are drawn at random without replacement. What is the probability that both are red?

Type a number. A table answer and a calculator answer are both accepted.

Show the answer

Multiplication rule: $P(\text{1st red}) \cdot P(\text{2nd red} \mid \text{1st red})$ $= \frac{5}{8} \cdot \frac{4}{7}$ $= \frac{20}{56}$ $\approx 0.357$ . After one red is gone, 4 of the 7 tokens left are red.

Answer: 0.3571

Part 9 · Summary

Summary

Given B, the outcomes in B are the new whole, so

P(A \mid B) = \frac{P(A \cap B)}{P(B)}

. Rearranged, the multiplication rule

P(A \cap B) = P(A) \cdot P(B \mid A)

is what a tree diagram does along each path. Adding paths gives a total probability, and dividing reverses a condition, which usually changes the answer.

Part 10 · Up next

What comes next

Part 11 · Connections