A food co-op surveyed 150 shoppers about whether they bring reusable bags and whether they buy in bulk. The results are below. | | Buys in bulk | No bulk | Total | |---|---|---|---| | Brings bags | 45 | 30 | 75 | | No bags | 15 | 60 | 75 | | Total | 60 | 90 | 150 | (a) Find the probability that a shopper brings reusable bags. (b) Find the probability that a shopper buys in bulk and brings bags. (c) Find the probability that a shopper buys in bulk given that they bring bags.

(a) P(bags) = 0.5; (b) P(bulk ∩ bags) = 0.3; (c) P(bulk | bags) = 0.6.

A coffee shop classified 200 drink orders by size and whether the drink was iced. | | Iced | Hot | Total | |---|---|---|---| | Large | 60 | 40 | 100 | | Small | 30 | 70 | 100 | | Total | 90 | 110 | 200 | Are the events 'the order is iced' and 'the order is large' independent? Justify your answer with a probability comparison.

Not independent: P(iced | large) = 0.60 but P(iced) = 0.45, so knowing an order is large changes the chance it is iced.

Conditional probability practice problems

By Jude Wallis · Published July 28, 2026

This set has 8 problems on reading two-way tables, finding the probability of A given B from the formula, checking whether two events are independent, and tree-style reasoning in words. Solve each on paper, then check your work against the full solution steps.

AP Statistics: Unit 2 (topics 2.2 Summary Statistics for Two Categorical Variables, 2.6 Conditional Probability, 2.7 Independent Events and Unions of Events). These problems cover Unit 2 topics 2.2 (conditional relative frequencies in two-way tables), 2.6 (conditional probability and the general multiplication rule), and 2.7 (independent events and the union of two events) in the Fall 2026 AP Statistics course. Unit 2 is 15 to 25% of the multiple-choice section.

What these problems build

These problems build four Unit 2 skills: reading joint, marginal, and conditional relative frequencies from a two-way table; applying the conditional probability formula $P(A \mid B) = \frac{P(A \cap B)}{P(B)}$ , where $P(A \mid B)$ is the probability of event $A$ given that event $B$ has happened and $P(A \cap B)$ is the probability that both happen; checking whether two events are independent; and reasoning through two-stage situations in words. Independent events satisfy $P(A \mid B) = P(A)$ , so knowing that $B$ happened does not change the chance of $A$ .

Work each problem on paper before opening the steps. For the difference between events that cannot happen together and events that do not influence each other, see disjoint vs independent events. Once you can read association from a two-way table, the chi-square calculator tests whether that association is strong enough to be statistically significant.

Problem 1

A food co-op surveyed 150 shoppers about whether they bring reusable bags and whether they buy in bulk. The results are below.

	Buys in bulk	No bulk	Total
Brings bags	45	30	75
No bags	15	60	75
Total	60	90	150

(a) Find the probability that a shopper brings reusable bags. (b) Find the probability that a shopper buys in bulk and brings bags. (c) Find the probability that a shopper buys in bulk given that they bring bags.

Show the worked solution

Read the totals you need from the table. There are 150 shoppers in all, 75 bring reusable bags, and 60 buy in bulk.
Part (a): the marginal probability that a shopper brings bags is the row total divided by the grand total, $P(\text{bags}) = \frac{75}{150} = 0.5$ .
Part (b): the joint probability that a shopper buys in bulk and brings bags is the cell where both meet divided by the grand total, $P(\text{bulk} \cap \text{bags}) = \frac{45}{150} = 0.3$ .
Part (c): the conditional probability restricts to shoppers who bring bags, so divide the 45 who do both by the 75 who bring bags, $P(\text{bulk} \mid \text{bags}) = \frac{45}{75} = 0.6$ .
Notice the denominator changed from 150 in part (b) to 75 in part (c). Conditioning on bringing bags shrinks the group you are looking at to that single row.

(a) $P(\text{bags}) = 0.5$ ; (b) $P(\text{bulk} \cap \text{bags}) = 0.3$ ; (c) $P(\text{bulk} \mid \text{bags}) = 0.6$ .

Problem 2

At a farmers market cheese stall, the probability that a customer samples the cheese is 0.45. The probability that a customer both samples the cheese and buys some is 0.18. (a) Find the probability that a customer buys cheese given that they sampled it. (b) Find the probability that a customer does not buy given that they sampled.

Show the worked solution

Write down what you are given: $P(\text{sample}) = 0.45$ and $P(\text{sample} \cap \text{buy}) = 0.18$ .
Part (a): apply the conditional probability formula with sampling as the condition, $P(\text{buy} \mid \text{sample}) = \frac{P(\text{sample} \cap \text{buy})}{P(\text{sample})} = \frac{0.18}{0.45}$ .
Divide: $\frac{0.18}{0.45} = 0.4$ , since $0.45 \times 0.4 = 0.18$ .
Part (b): given that a customer sampled, buying and not buying are the only two options, so their conditional probabilities add to 1.
Subtract: $P(\text{no buy} \mid \text{sample}) = 1 - 0.4 = 0.6$ .

(a) $P(\text{buy} \mid \text{sample}) = 0.4$ ; (b) $P(\text{no buy} \mid \text{sample}) = 0.6$ .

Problem 3

A coffee shop classified 200 drink orders by size and whether the drink was iced.

	Iced	Hot	Total
Large	60	40	100
Small	30	70	100
Total	90	110	200

Are the events 'the order is iced' and 'the order is large' independent? Justify your answer with a probability comparison.

Show the worked solution

Find the marginal probability that an order is iced by dividing the iced column total by the grand total, $P(\text{iced}) = \frac{90}{200} = 0.45$ .
Find the conditional probability that an order is iced given it is large by restricting to the 100 large orders, $P(\text{iced} \mid \text{large}) = \frac{60}{100} = 0.60$ .
Compare the two. Independence would require $P(\text{iced} \mid \text{large}) = P(\text{iced})$ , but $0.60 \neq 0.45$ .
As a second check, compare the joint probability with the product of the marginals. The joint is $P(\text{iced} \cap \text{large}) = \frac{60}{200} = 0.30$ .
The product is $P(\text{iced}) \times P(\text{large}) = 0.45 \times \frac{100}{200} = 0.45 \times 0.5 = 0.225$ .
Because $0.30 \neq 0.225$ , both tests agree that size and being iced are not independent.

Not independent: $P(\text{iced} \mid \text{large}) = 0.60$ but $P(\text{iced}) = 0.45$ , so knowing an order is large changes the chance it is iced.

Problem 4

A phone-charging kiosk holds 10 cables: 4 are USB-C and 6 are micro-USB. Two travelers each take one cable at random, one after another, without replacement. (a) Find the probability that both take a USB-C cable. (b) Find the probability that the first traveler takes a micro-USB cable and the second takes a USB-C cable.

Show the worked solution

Part (a): the first traveler faces 4 USB-C cables out of 10, so $P(\text{first USB-C}) = \frac{4}{10}$ .
After one USB-C cable is gone, 3 USB-C cables remain out of 9 cables, so $P(\text{second USB-C} \mid \text{first USB-C}) = \frac{3}{9}$ .
By the general multiplication rule, multiply along the branch: $P(\text{both USB-C}) = \frac{4}{10} \times \frac{3}{9} = \frac{12}{90} = \frac{2}{15} \approx 0.133$ .
Part (b): the first traveler faces 6 micro-USB cables out of 10, so $P(\text{first micro}) = \frac{6}{10}$ .
With one micro-USB cable gone, all 4 USB-C cables remain out of 9, so $P(\text{second USB-C} \mid \text{first micro}) = \frac{4}{9}$ .
Multiply along the branch: $P(\text{micro then USB-C}) = \frac{6}{10} \times \frac{4}{9} = \frac{24}{90} = \frac{4}{15} \approx 0.267$ .

(a) $\frac{2}{15} \approx 0.133$ ; (b) $\frac{4}{15} \approx 0.267$ .

Problem 5

A car wash logged 300 vehicles by type and whether the driver added the wax upgrade.

	Wax	No wax	Total
SUV	60	60	120
Not SUV	40	140	180
Total	100	200	300

(a) Find the probability a vehicle added wax given it is an SUV. (b) Find the probability a vehicle is an SUV given it added wax. (c) Explain why the two answers differ.

Show the worked solution

Part (a): restrict to the 120 SUVs and count how many added wax, $P(\text{wax} \mid \text{SUV}) = \frac{60}{120} = 0.50$ .
Part (b): now restrict to the 100 vehicles that added wax and count how many were SUVs, $P(\text{SUV} \mid \text{wax}) = \frac{60}{100} = 0.60$ .
Both parts start from the same 60 vehicles in the SUV-and-wax cell, but the denominators differ.
In part (a) you divide by 120 (all SUVs); in part (b) you divide by 100 (all wax buyers), which is why $0.50 \neq 0.60$ .
Part (c): the two conditionals answer different questions. $P(\text{wax} \mid \text{SUV})$ asks what fraction of SUVs added wax, while $P(\text{SUV} \mid \text{wax})$ asks what fraction of wax buyers drove SUVs, so swapping the condition changes the result.

(a) $P(\text{wax} \mid \text{SUV}) = 0.50$ ; (b) $P(\text{SUV} \mid \text{wax}) = 0.60$ ; (c) they differ because the condition, and so the denominator, changes from all SUVs to all wax buyers.

Problem 6

A bike shop asked 240 customers whether they took a test ride and whether they bought a bike. In all, 150 customers took a test ride and 90 customers bought a bike. Among the customers who took a test ride, 72 bought a bike. (a) Complete the two-way table. (b) Find the probability that a customer bought a bike given that they did not take a test ride. (c) Are buying a bike and taking a test ride independent? Justify with a calculation.

Show the worked solution

Start with what you know: 240 customers total, 150 took a test ride, 90 bought a bike, and 72 both test-rode and bought.
Fill the test-ride row: if 72 of the 150 test riders bought, then $150 - 72 = 78$ test rode but did not buy.
Find the no-test-ride totals: $240 - 150 = 90$ customers did not test ride, and $90 - 72 = 18$ of the 90 buyers did not test ride.
Complete the last cell: $90 - 18 = 72$ customers neither test rode nor bought. The finished table is:
Bought Not bought Total
Test ride 72 78 150
No test ride 18 72 90
Total 90 150 240
Part (b): restrict to the 90 customers who did not test ride, of whom 18 bought, $P(\text{bought} \mid \text{no test ride}) = \frac{18}{90} = 0.2$ .
Part (c): the overall buy rate is $P(\text{bought}) = \frac{90}{240} = 0.375$ , while among test riders it is $P(\text{bought} \mid \text{test ride}) = \frac{72}{150} = 0.48$ .
Because $0.48 \neq 0.375$ , the two events are not independent; test riders bought at a higher rate than customers overall.

	Bought	Not bought	Total
Test ride	72	78	150
No test ride	18	72	90
Total	90	150	240

(a) test ride: 72 bought, 78 not; no test ride: 18 bought, 72 not; (b) $P(\text{bought} \mid \text{no test ride}) = 0.2$ ; (c) not independent, since $P(\text{bought} \mid \text{test ride}) = 0.48 \neq 0.375 = P(\text{bought})$ .

Problem 7

A bakery bakes croissants in two ovens. Oven A produces 60% of the croissants and oven B produces the other 40%. Of oven A's croissants, 5% come out underbaked; of oven B's, 10% come out underbaked. (a) Find the probability that a randomly chosen croissant is underbaked. (b) Given that a croissant is underbaked, find the probability it came from oven B.

Show the worked solution

Set up a hypothetical batch of 1000 croissants so the percentages become counts. Oven A makes $0.60 \times 1000 = 600$ and oven B makes $0.40 \times 1000 = 400$ .
Count oven A's underbaked croissants: $0.05 \times 600 = 30$ .
Count oven B's underbaked croissants: $0.10 \times 400 = 40$ .
Part (a): add the two underbaked counts and divide by the batch size, $P(\text{underbaked}) = \frac{30 + 40}{1000} = \frac{70}{1000} = 0.07$ .
Part (b): restrict to the 70 underbaked croissants and find how many came from oven B, $P(\text{oven B} \mid \text{underbaked}) = \frac{40}{70} \approx 0.571$ .
The formula gives the same value: $P(\text{oven B} \mid \text{underbaked}) = \frac{P(\text{oven B} \cap \text{underbaked})}{P(\text{underbaked})} = \frac{0.40 \times 0.10}{0.07} = \frac{0.04}{0.07} \approx 0.571$ .

(a) $P(\text{underbaked}) = 0.07$ ; (b) $P(\text{oven B} \mid \text{underbaked}) = \frac{40}{70} \approx 0.571$ .

Problem 8

In a trivia app, the probability a player answers the sports question correctly is 0.7 and the probability they answer the history question correctly is 0.6. The two questions are independent. (a) Find the probability the player gets both correct. (b) Find the probability they get at least one correct. (c) Find the probability they get exactly one correct. (d) Given that they get at least one correct, find the probability they get both correct.

Show the worked solution

Part (a): because the questions are independent, multiply the two probabilities, $P(\text{both}) = 0.7 \times 0.6 = 0.42$ .
Part (b): use the union rule, $P(\text{at least one}) = P(\text{sports}) + P(\text{history}) - P(\text{both}) = 0.7 + 0.6 - 0.42 = 0.88$ .
Part (c): exactly one correct is the union with the both-correct overlap removed, $P(\text{exactly one}) = P(\text{at least one}) - P(\text{both}) = 0.88 - 0.42 = 0.46$ .
Check part (c) by adding the two one-correct cases: sports only is $0.7 \times (1 - 0.6) = 0.7 \times 0.4 = 0.28$ , history only is $(1 - 0.7) \times 0.6 = 0.3 \times 0.6 = 0.18$ , and $0.28 + 0.18 = 0.46$ .
Part (d): getting both correct already guarantees at least one is correct, so $P(\text{both} \cap \text{at least one}) = P(\text{both}) = 0.42$ .
Divide by the probability of the condition, $P(\text{both} \mid \text{at least one}) = \frac{0.42}{0.88} \approx 0.477$ .

(a) 0.42; (b) 0.88; (c) 0.46; (d) $P(\text{both} \mid \text{at least one}) = \frac{0.42}{0.88} \approx 0.477$ .