2.6 - Conditional Probability

Overview

A conditional probability is a probability that uses extra information: you already know one event happened, so you only look at the outcomes where it did. This topic covers the conditional probability formula and the general multiplication rule, which turns that formula around to find the probability that two events both happen. These are the tools behind tree diagrams, reading medical test results, and checking independence in Topic 2.7.

What "given" means

P(A \mid B)

is read "the probability of

A

given

B

." It's the probability that

A

happens when you know that

B

has happened. The vertical bar means given, and the event after the bar is the condition.

Knowing that

B

happened shrinks the sample space. Every outcome outside

B

is ruled out, and

B

becomes the new whole. So

P(A \mid B)

asks: out of the outcomes in

B

, what fraction are also in

A

In a two-way table, this means you stop using the grand total and use only the row or column for the condition. That's the conditional relative frequency from Topic 2.2.

Example 1: Conditional probability from a two-way table

Here again is the survey of all 400 seniors at a high school from Topic 2.5.

	Job	No job	Total
License	126	114	240
No license	34	126	160
Total	160	240	400

One senior is chosen at random. Let

L

= "has a license" and

J

= "has a job."

(a) Find

P(J \mid L)

Step 1: The condition is

L

, so restrict to the License row. It has 240 seniors.

Step 2: Of those 240 seniors, 126 have a job.

P(J \mid L) = \frac{126}{240} = 0.525

Given that a senior has a license, the probability that they have a part-time job is

0.525

. So 52.5% of seniors with a license have a job.

(b) Find

P(L \mid J)

Step 1: Now the condition is

J

, so restrict to the Job column. It has 160 seniors.

Step 2: Of those 160 seniors, 126 have a license.

P(L \mid J) = \frac{126}{160} = 0.7875

Given that a senior has a part-time job, the probability that they have a license is about

0.788

. About 78.8% of seniors with a job have a license.

Notice:

P(J \mid L)

and

P(L \mid J)

use the same cell (126) but different totals, so they're different. Order matters: the event after the bar decides which total you divide by.

The conditional probability formula

When you have probabilities instead of counts, use the formula from the AP formula sheet:

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

This works whenever

P(B) > 0

(you can't condition on something impossible). It's the same idea as the table: the joint probability is the part of

B

that's also in

A

, and

P(B)

is the new whole. Example 1 gives the same answer with the formula:

P(J \mid L) = \frac{P(J \cap L)}{P(L)} = \frac{126/400}{240/400} = \frac{126}{240} = 0.525

The complement rule from Topic 2.4 still works inside a condition:

P(A^C \mid B) = 1 - P(A \mid B)

. Once you know

B

happened,

A

either happens or it doesn't.

Example 2: Using the formula

At a large university, 60% of students live on campus, 55% have a meal plan, and 48% live on campus and have a meal plan. A student is chosen at random. Let

C

= "lives on campus" and

M

= "has a meal plan."

(a) Find the probability that the student has a meal plan, given that they live on campus.

P(M \mid C) = \frac{P(M \cap C)}{P(C)} = \frac{0.48}{0.60} = 0.80

Among students who live on campus, 80% have a meal plan.

(b) Find the probability that the student does not have a meal plan, given that they live on campus.

P(M^C \mid C) = 1 - P(M \mid C) = 1 - 0.80 = 0.20

Among students who live on campus, 20% don't have a meal plan.

(c) Find the probability that the student lives on campus, given that they have a meal plan.

P(C \mid M) = \frac{P(C \cap M)}{P(M)} = \frac{0.48}{0.55} \approx 0.873

Among students with a meal plan, about 87.3% live on campus. Again,

P(C \mid M) \neq P(M \mid C)

, because the condition changes the denominator.

The general multiplication rule

Write the formula with

A

as the condition,

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

, and multiply both sides by

P(A)

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

This is the general multiplication rule. In words: the probability that both events happen equals the probability that the first one happens, times the probability that the second one happens given that the first one did. It works for any two events, which is why it's called general. (In Topic 2.7, you'll see the shortcut for independent events, where

P(B \mid A)

is just

P(B)

Use it when events happen in a sequence and the first result affects the second, like choosing people or objects without replacement.

Example 3: Choosing without replacement

A class has 12 juniors and 8 seniors. The teacher randomly picks one student to present first, then randomly picks a different student to present second. Find the probability that both presenters are seniors.

Step 1: Find the probability that the first pick is a senior. There are 8 seniors out of 20 students.

P(\text{1st senior}) = \frac{8}{20}

Step 2: Find the probability that the second pick is a senior, given that the first one was. Now there are 7 seniors left out of 19 students.

P(\text{2nd senior} \mid \text{1st senior}) = \frac{7}{19}

Step 3: Multiply.

P(\text{both seniors}) = \frac{8}{20} \cdot \frac{7}{19} = \frac{56}{380} \approx 0.147

Answer: The probability that both presenters are seniors is about

0.147

Tree diagrams

When a random process happens in stages, a tree diagram keeps the conditional probabilities organized.

The first set of branches shows the first stage. Its probabilities add to 1.

Each later set of branches shows the next stage, given the branch it grows from. The branches coming out of each point add to 1.

Multiply along a path to get the probability of that whole path. That's the general multiplication rule.

Add the paths that lead to the event you want. Different paths can't both happen (they're mutually exclusive), so their probabilities add.

Example 4: A screening test (AP-style)

A disease affects 2% of a large population. A screening test gives a positive result for 95% of people who have the disease. It also gives a positive result for 8% of people who don't have the disease (a false positive). A person is chosen at random and tested. Let

D

= "has the disease" and

+

= "tests positive."

(a) Find the probability that the person tests positive.

(b) Given that the person tests positive, find the probability that they have the disease.

Step 1: Draw the tree. The first stage is disease status. The second stage is the test result, given disease status.

Step 2: Find each path that ends in a positive test (in red) with the general multiplication rule.

P(D \cap +) = P(D) \cdot P(+ \mid D) = 0.02(0.95) = 0.019

P(D^C \cap +) = P(D^C) \cdot P(+ \mid D^C) = 0.98(0.08) = 0.0784

Step 3 (a): A positive test can happen on either red path, so add them.

P(+) = 0.019 + 0.0784 = 0.0974

The probability that a randomly chosen person tests positive is

0.0974

Step 4 (b): The tree gives

P(+ \mid D)

, but the question asks for

P(D \mid +)

. Use the conditional probability formula.

P(D \mid +) = \frac{P(D \cap +)}{P(+)} = \frac{0.019}{0.0974} \approx 0.195

Answer: Given a positive test, the probability that the person has the disease is only about

0.195

. Fewer than 1 in 5 people who test positive actually have the disease.

Why so low? Picture 10,000 people. About 200 have the disease, and 190 of them test positive. The other 9,800 don't have it, but 8% of them (784 people) also test positive. Out of

190 + 784 = 974

positive tests, only 190 are true positives, and

\frac{190}{974} \approx 0.195

. When a condition is rare, false positives from the large healthy group can outnumber the true positives.

Showing your work on the AP exam

On free-response questions, a probability calculation earns full credit when it shows three things:

The formula or structure, with the events named, for example $P(D \mid +) = \frac{P(D \cap +)}{P(+)}$ .

The numbers from the problem substituted in.

The answer, interpreted in context.

A bare number, or a calculator result with no setup, can lose credit even when it's correct.

Common mistakes

Dividing by the grand total. A conditional probability divides by the total for the condition (its row, its column, or $P(B)$ ), not by the whole table.

Flipping the condition. $P(+ \mid D) = 0.95$ is not the same as $P(D \mid +) \approx 0.195$ . Read which event comes after the bar.

Putting the wrong event in the denominator. In $P(A \mid B)$ , you divide by $P(B)$ , the probability of the event after the bar.

Adding along a path or multiplying across paths. Multiply along a single path, and add separate paths.

Forgetting that the second draw changes. Without replacement, the second probability is conditional: $\frac{7}{19}$ , not $\frac{8}{20}$ .

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