Probability Theory Workbook: 10 Exercises with Full Solutions

This workbook accompanies the Probability Theory lesson and is built to be worked through with a pen and paper. The ten exercises move in the same order as the lesson, starting from equally likely outcomes and building up through the additive law, independence, conditional probability, and finally Bayes’ theorem with probability trees. Each one is designed to test a single idea cleanly, so when you get stuck you know exactly which concept to revisit.

The recommended way to use it is to attempt all ten exercises first, writing out your reasoning rather than just the final number, and only then turn to the worked solutions. The solutions show every step and name the rule being applied, so the goal is not just to check your answer but to compare your method against the standard one. A correct answer reached by luck is worth less than a clear method that you can repeat on the exam.

Part One: The Exercises

Exercise 1 (Equally likely outcomes). A single card is drawn at random from a standard deck of 52 cards. Find the probability that the card is a heart, and the probability that the card is a face card (jack, queen, or king).

Exercise 2 (The axioms). A weather service proposes the following forecast for tomorrow, where the three outcomes are mutually exclusive and collectively exhaustive: P(sunny) = 0.5, P(cloudy) = 0.3, P(rainy) = 0.2. A second service proposes P(sunny) = 0.6, P(cloudy) = 0.5, P(rainy) = 0.1. Using the axioms of probability, decide which forecast is valid and explain why the other is not.

Exercise 3 (Additive law with overlap). In a sports club, 60% of members play football, 30% play basketball, and 20% play both. Find the probability that a randomly chosen member plays at least one of the two sports.

Exercise 4 (Complement rule). A fair six-sided die is rolled twice. Using the complement rule, find the probability of getting at least one six across the two rolls.

Exercise 5 (Mutually exclusive events). A single card is drawn from a standard deck. Find the probability that it is either a king or a queen, and state why the additive law simplifies in this case.

Exercise 6 (Independence). Two events A and B are independent, with P(A) = 0.4 and P(B) = 0.5. Find the probability that both occur, and the probability that at least one occurs.

Exercise 7 (Conditional probability). In a survey, 50% of respondents own a bicycle, and 20% of all respondents own both a bicycle and a car. Given that a respondent owns a bicycle, find the probability that they also own a car.

Exercise 8 (Multiplicative law, without replacement). An urn contains 5 red balls and 3 blue balls. Two balls are drawn one after the other without replacement. Find the probability that both balls are red.

Exercise 9 (Bayes’ theorem). A disease affects 1% of a population. A diagnostic test correctly returns positive for 99% of people who have the disease, and incorrectly returns positive for 5% of people who do not. A person tests positive. Find the probability that they actually have the disease.

Exercise 10 (Total probability and a probability tree). A company sources a component from two factories. Factory 1 supplies 60% of the components and has a 2% defect rate. Factory 2 supplies the remaining 40% and has a 5% defect rate. Find the overall probability that a randomly selected component is defective, and then, given that a component is defective, find the probability that it came from Factory 2.

Part Two: Worked Solutions (please try and solve them by yourself first)

Solution 1. With equally likely outcomes, probability is the number of favourable outcomes divided by the total number of outcomes.

There are 13 hearts among the 52 cards, giving one quarter. For face cards there are three per suit across four suits, so twelve in total.

Solution 2. The axioms require every probability to be non-negative, and for mutually exclusive and collectively exhaustive outcomes the probabilities must sum to exactly 1, since one of them is certain to occur. Checking the first forecast:

The first forecast is valid because the outcomes are non-negative and sum to one. Checking the second:

The second forecast is invalid, because a set of mutually exclusive and exhaustive outcomes cannot have probabilities that sum to more than 1.

Solution 3. Because some members play both sports, the two events overlap, so the additive law subtracts the intersection to avoid double counting.

There is a 70% chance a randomly chosen member plays at least one of the two sports.

Solution 4. The event “at least one six” is awkward to count directly, so the complement rule turns it into the easier event “no six at all.” On a single roll the probability of not getting a six is 5/6, and the two rolls are independent.

The probability is 11/36, or about 0.306.

Solution 5. A single card cannot be both a king and a queen at the same time, so the two events are mutually exclusive and their intersection is zero. The additive law therefore reduces to a simple sum.

The probability is 2/13.

Solution 6. For independent events, the probability of both occurring is the product of the individual probabilities.

To find the probability that at least one occurs, apply the additive law using this intersection.

Both occur with probability 0.2, and at least one occurs with probability 0.7.

Solution 7. Conditional probability rescales the intersection by the probability of the condition.

Among bicycle owners, 40% also own a car.

Solution 8. The two draws are dependent, because removing the first ball changes what remains for the second. The multiplicative law combines the first draw with the second draw conditioned on the first.

The probability that both balls are red is 5/14, or about 0.357.

Solution 9. Let D be the event of having the disease and “+” the event of a positive test. The known values are P(D) = 0.01, P(+ | D) = 0.99, and P(+ | not D) = 0.05. First find the overall probability of a positive test using the total probability formula.

Then apply Bayes’ theorem.

Despite the test being highly accurate, a positive result means only about a 17% chance of actually having the disease, because the disease is rare and false positives outnumber true positives.

Solution 10. Let F1 and F2 be the events that the component came from each factory, and D the event that it is defective. The known values are P(F1) = 0.60, P(F2) = 0.40, P(D | F1) = 0.02, and P(D | F2) = 0.05. The overall defect rate follows from the total probability formula, summing the two paths of the probability tree that end in a defect.

So 3.2% of components are defective overall. To find the source of a defective component, apply Bayes’ theorem, dividing the Factory 2 path by the total.

Even though Factory 2 supplies fewer components, its higher defect rate means that 62.5% of all defective components trace back to it.

How to Get the Most From This Workbook

Notice the pattern that runs through the harder exercises. Whenever a problem involves stages, a first draw then a second, a factory then a defect, a true state then a test result, it can be drawn as a probability tree, and the answer comes from multiplying along the branches of a path and summing across the paths that lead to the outcome you care about. Exercises 8, 9, and 10 are all the same machinery in different clothing, which is exactly the realisation that makes the exam version feel routine. Work them until that structure is obvious, then move on to the next lesson with the confidence that the foundations are solid.

See you soon.