Discrete Probability Distributions: 10 Exercises with Full Solutions

This workbook accompanies the Discrete Probability Distributions article and is built for practice with a calculator. The ten exercises move through the same family as the lesson: the uniform, binomial, Poisson, geometric, and negative binomial distributions, together with expected value, variance, and the rules for linear combinations. Each one isolates a single technique so you can see exactly where your method needs work.

Attempt every exercise before reading the solutions, and write out each step. The most reliable habit in this topic is to identify the distribution first, the number of trials, whether you are counting successes or waiting for them, the rate, before reaching for a formula, because choosing the right distribution is most of the battle. Where an exponential value is needed, it is supplied in the solution.

Part One: The Exercises

Exercise 1 (Discrete uniform). A fair six-sided die is rolled. Find the probability that the score is at least 5, and state the mean of the distribution.

Exercise 2 (Binomial probability). A basketball player makes each free throw independently with probability 0.3. In five attempts, find the probability of making exactly two.

Exercise 3 (Binomial via the complement). A fair coin is tossed six times. Find the probability of getting at least one head.

Exercise 4 (Binomial mean and variance). A quality inspector checks 20 independent items, each defective with probability 0.1. Find the expected number of defectives, the variance, and the standard deviation.

Exercise 5 (Poisson probability). A call centre receives calls at an average rate of 2.5 per hour. Find the probability of receiving exactly three calls in an hour.

Exercise 6 (Scaling the Poisson rate). For the same call centre, find the probability of receiving no calls in a 30-minute period.

Exercise 7 (Poisson approximation to the binomial). A factory produces components with a 1% defect rate. In a batch of 200, use the Poisson approximation to find the probability of exactly one defective, and confirm the approximation conditions hold.

Exercise 8 (Expected value and variance from a table). A discrete random variable X takes the values 0, 1, 2, and 3 with probabilities 0.1, 0.3, 0.4, and 0.2. Compute the expected value and the variance.

Exercise 9 (Linear combination). Using the X from Exercise 8, define Y = 5 + 10X. Find the expected value and variance of Y.

Exercise 10 (Geometric and negative binomial). A game is won on each independent attempt with probability 0.25. (a) Find the probability that the first win occurs on the third attempt, and state the expected number of attempts until the first win. (b) Find the probability that the second win occurs on the fifth attempt.

Part Two: Worked Solutions

Solution 1. On a fair die each score has probability one sixth, so the probability of at least 5 is the sum of the probabilities of 5 and 6.

The mean of a discrete uniform over k values is (k + 1) over 2, so for k equal to 6 it is 3.5.

Solution 2. The number of successes is binomial with n equal to 5 and pi equal to 0.3, so the probability of exactly two uses the binomial formula.

Solution 3. The number of heads is binomial with n equal to 6 and pi equal to 0.5, and “at least one” is found through the complement.

Solution 4. With n equal to 20 and pi equal to 0.1, the binomial mean and variance follow directly.

The standard deviation is the square root of the variance, about 1.34.

Solution 5. The number of calls in an hour is Poisson with rate 2.5, so the probability of exactly three uses the Poisson formula, with the exponential value about 0.0821.

Solution 6. A 30-minute period is half a unit, so the rate scales to 1.25, and the probability of zero calls is the Poisson formula at zero, which reduces to the exponential of the negative rate.

Solution 7. The Poisson approximation sets lambda equal to n times pi, which is 200 times 0.01, or 2. The conditions hold: n of 200 exceeds 30, n times pi of 2 is below 10, and the value 1 is small relative to 200. The probability of exactly one is then the Poisson formula at one.

Solution 8. The expected value is the probability-weighted sum of the values.

For the variance, first find the expected value of the square, then subtract the squared mean.

Solution 9. Expectation is linear, so the constant shifts the mean and the multiplier scales it.

Variance ignores the additive constant and scales by the square of the multiplier.

The constant 5 vanishes from the variance because adding a constant shifts the distribution without changing its spread.

Solution 10. In part (a), the trial of the first success is geometric, so the first win on the third attempt requires two failures followed by a win.

The expected number of attempts until the first win is one over the success probability.

In part (b), the trial of the second success is negative binomial with r equal to 2, so the second win on the fifth attempt needs one win among the first four trials and a win on the fifth.

How to Get the Most From This Workbook

The single decision that drives every one of these problems is identifying the distribution before touching a formula. Ask whether the trials are fixed in number, in which case a count of successes is binomial, or whether you are waiting for successes, in which case the geometric handles the first and the negative binomial the rth. Reach for the Poisson when events occur at a rate in a continuous medium, remembering to scale the rate to the period you care about, and use it as a shortcut for the binomial when the number of trials is large and successes are rare. For any distribution, the mean and variance come straight from the reference table once you know which one you are in, and the linear-combination rules let you handle any rescaled version: the mean is linear while the variance scales by the square of the multiplier and drops additive constants. Drill the identification step until it is automatic, and the arithmetic that follows becomes routine.

See you soon.