Unit 2 · Topic 2.6 · about 25 minutes

Conditional Probability

Calculate a conditional probability from a table or from given probabilities, and use the general multiplication rule to find the probability that two events both happen.

Predict first

A sports league tests its athletes for a banned substance. Suppose 4% of the athletes use it. The test comes back positive for 95% of users, and also, wrongly, for 6% of athletes who do not use it. An athlete tests positive. About how likely is it that this athlete uses the substance?

What "given" does

A conditional probability is the probability that event A occurs given that event B has occurred. It is written $P (A | B)$ and read "the probability of A given B". Knowing that B happened shrinks the possibilities to the outcomes in B, and the question becomes what fraction of those are also in A:

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

This is the probability version of the conditional relative frequency from 2.2 Summary Statistics for Two Categorical Variables. With a table of counts, "given" tells you which row or column to stay inside, and that row or column total becomes the denominator.

600 juniors and seniors by part-time job (counts)
	Has a job	No job	Total
Junior	96	224	320
Senior	140	140	280
Total	236	364	600

Choose one of these 600 students at random. Given that the student is a senior, you stay in the senior row, so $P (job | senior) = \frac{140}{280} = 0.50$ . The formula agrees: $\frac{P (job \cap senior)}{P (senior)} = \frac{140 / 600}{280 / 600} = 0.50$ .

Now reverse the condition. Given that the student has a job, you stay in the job column: $P (senior | job) = \frac{140}{236} \approx 0.593$ . The numerator is the same, but the denominator changes because the question changed. Half of the seniors have jobs, and about 59% of the students with jobs are seniors.

The general multiplication rule

Write the definition for $P (B | A)$ instead, multiply both sides by $P (A)$ , and you get the general multiplication rule:

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

For A and B both to happen, A has to happen, and then B has to happen given that A did. It is the tool to reach for when events happen in sequence and the second depends on the first.

A club has 12 members, 5 of them seniors. Two members are chosen at random as president and treasurer, and nobody can hold both jobs. The chance that the president is a senior is $\frac{5}{12}$ . Given that a senior was chosen, 4 of the remaining 11 members are seniors, so

P (both seniors) = \frac{5}{12} \cdot \frac{4}{11} = \frac{20}{132} \approx 0.152

Worked exampleBack to the drug test

In the league from the prediction, 4% of athletes use the banned substance. The test is positive for 95% of users and for 6% of non-users. Find the probability that an athlete who tests positive is a user.

Name the events and sort the facts. Let U be "uses the substance" and + be "tests positive". The facts are $P (U) = 0.04$ , $P (+ | U) = 0.95$ and $P (+ | U^{C}) = 0.06$ . The question asks for $P (U | +)$ , with the condition on the other side.
Find each way to test positive with the multiplication rule. User and positive: $P (U \cap +) = P (U) \cdot P (+ | U) = 0.04 (0.95) = 0.038$ . Non-user and positive: $P (U^{C} \cap +) = 0.96 (0.06) = 0.0576$ .
Add them for the total chance of a positive. $P (+) = 0.038 + 0.0576 = 0.0956$ .
Apply the definition. $P (U | +) = \frac{P (U \cap +)}{P (+)} = \frac{0.038}{0.0956} \approx 0.397$ .
Interpret in context. About 39.7% of the athletes who test positive actually use the substance. Most positive tests are false alarms, because non-users far outnumber users.

Answer.

$P (U | +) \approx 0.397$ . A positive test raises the chance of use from 0.04 to about 0.40, but on its own it is not enough to conclude that this athlete uses the substance.

The same probabilities as counts for 10,000 imagined athletes. The answer is 380 out of 956 positives.
	Tests positive	Tests negative	Total
User	380	20	400
Non-user	576	9,024	9,600
Total	956	9,044	10,000

Check your understanding

Of 320 juniors at a school, 96 have a part-time job. Of 280 seniors, 140 have one. One of these 600 students is chosen at random. Given that the student has no job, what is the probability that the student is a junior? Round to three decimal places.

Your answer

A survey found that among adults who own a dog, 70% walk at least 30 minutes a day. Let D be "owns a dog" and W be "walks at least 30 minutes a day" for a randomly chosen adult. Which statement matches the survey result?

A box holds 20 phone chargers, and 3 of them are defective. Two chargers are taken out at random, one after the other, without replacement. What is the probability that both are defective?

At a coffee shop, 60% of customers order a hot drink. Of the customers who order a hot drink, 25% also buy a pastry. What is the probability that a randomly chosen customer orders a hot drink and buys a pastry?

For a randomly chosen adult in a town, the probability of having a library card is 0.40, and the probability of having a library card and visiting the library in the past month is 0.12. Given that an adult has a library card, what is the probability that the adult visited the library in the past month?

Practice

Practice until it is automatic

Work the problems that say "given" first: find the group you are restricted to, and use its total as the denominator. The rest read a probability straight from a table or use the complement rule from 2.4 Introduction to Probability.

Probability from two-way tables practice page · Addition rule, complements and conditional probability practice page

Course alignment, for teachers

AP Statistics topic 2.6, Unit 2: Probability, Random Variables, and Probability Distributions.

Skill 3.C: Calculate and estimate expected counts, percentages, probabilities, and intervals.