Source: Practice Exam 1, Spring 2016 | Purdue University
Difficulty: Intermediate | Prerequisites: Descriptive statistics notes, basic integration (for continuous distributions).
Probability distributions are the bridge between raw data and inference. Once you can model a random variable with the right distribution (binomial, geometric, uniform, Poisson, etc.), you unlock formulas for its expected value, variance and probabilities. This section is heavily tested: the exam will give you a scenario and expect you to identify the correct distribution, set up the parameters, and compute probabilities and expectations. You need to be comfortable with both discrete distributions (binomial) and continuous ones (uniform), as well as the algebra of expectations.
Every probability question starts with identifying the right distribution from the scenario. For discrete trials with two outcomes and a fixed count, use binomial. For a continuous variable equally likely across an interval, use uniform. Once you have the distribution, expected value and probability calculations follow from its formulas. The linearity of expectation (E[aX + bY] = aE[X] + bE[Y]) works regardless of independence.
Binomial distribution
Models the number of successes in n independent trials, each with the same probability of success p. In simple terms, "how many times does X happen in n tries?"
Geometric distribution
Models the number of trials needed to get the first success. In simple terms, "how long until it happens for the first time?" The key phrase in exam scenarios is "number of trials until the first..."
Poisson distribution
Models the number of events occurring in a fixed interval of time or space, given a known average rate. Think of it as counting rare, independent events.
Continuous uniform distribution
A distribution where every value in an interval [a, b] is equally likely. The PDF is a flat line at height 1/(b - a) over that interval and zero elsewhere.
Probability density function (PDF)
For a continuous random variable, the function f(x) whose integral over an interval gives the probability that the variable falls in that interval. The total area under the PDF is always 1.
Expected value (E[X])
The long-run average of a random variable. For a discrete distribution, it is the sum of each value times its probability. For a continuous distribution, it is the integral of x times f(x).
E[g(X)]
The expected value of a function of X. Computed as the integral (or sum) of g(x) times f(x). A common exam question asks for E[X²], where g(x) = x².
Conditional probability
The probability of event A given that event B has occurred: P(A|B) = P(A and B) / P(B). In simple terms, you are restricting the sample space to only those outcomes where B happened.
Linearity of expectation
E[aX + bY] = aE[X] + bE[Y]. This holds whether or not X and Y are independent. It is the workhorse for combining expected values.
Fixed number of independent trials, two outcomes (success/failure), same probability each trial: Binomial.
Example: rolling a die 5 times and counting how many 3s you get. n = 5, p = 1/8 for an 8-sided die.
Counting trials until the first success: Geometric.
Example: number of matches until the first blue-uniform win.
Note: the exam practice key marked this as "none of the above" because the scenario described "number of matches until the first competitor with a blue uniform won," which is geometric. If geometric is not among the choices, the answer is "none of the above."
Counting events in a fixed interval with a known average rate: Poisson.
Every value in an interval equally likely (continuous): Uniform.
P(X = k) = C(n, k) × p^k × (1 - p)^(n - k)
For "more than" questions, use the complement: P(X > 1) = 1 - P(X = 0) - P(X = 1).
Example: n = 5, p = 0.125. P(X > 1) = 1 - C(5,0)(0.125)⁰(0.875)⁵ - C(5,1)(0.125)¹(0.875)⁴ = 1 - 0.513 - 0.366 = 0.121.
P(X = 2 | X > 1) = P(X = 2 and X > 1) / P(X > 1) = P(X = 2) / P(X > 1).
The numerator simplifies because if X = 2 then X > 1 is automatically true.
Example: P(X = 2) = C(5,2)(0.125)²(0.875)³ = 0.107. Then 0.107 / 0.121 = 0.884.
For a function g(X): E[g(X)] = integral of g(x) × f(x) dx over the support.
Example with uniform on [0, 2] and g(x) = x²: E[X²] = integral from 0 to 2 of x² × (1/2) dx = (1/2) × [x³/3] from 0 to 2 = (1/2)(8/3) = 4/3 = 1.33.
Key steps the exam grades on: writing the definition, substituting limits and f(x), performing the integration, evaluating, and simplifying.
For a binomial: E[X] = np.
Example: n = 5, p = 0.125, so E[X] = 0.625.
Linearity of expectation: E[aX + bY] = aE[X] + bE[Y].
Example: E[2X - 3Y] = 2(0.625) - 3(0.8) = 1.25 - 2.4 = -1.15.
Independence is not required for linearity to hold.
Binomial PMF:
P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}Binomial expected value:
E[X] = npContinuous uniform PDF on [a, b]:
f_X(x) = \frac{1}{b - a}, \quad a \leq x \leq bExpected value of g(X) for a continuous variable:
E[g(X)] = \int_{-\infty}^{\infty} g(x) \, f(x) \, dxConditional probability:
P(A \mid B) = \frac{P(A \cap B)}{P(B)}Linearity of expectation:
E[aX + bY] = a\,E[X] + b\,E[Y]Students confuse geometric and binomial scenarios. Binomial counts successes in a fixed number of trials. Geometric counts trials until the first success. If the question says "how many trials until," it is geometric.
When computing E[X²] for a continuous variable, students sometimes write E[X]² instead. These are different quantities. E[X²] requires integrating x² × f(x); E[X]² is just the square of the mean.
Students forget to use the complement rule for "more than" probabilities. P(X > 1) is easier as 1 - P(X = 0) - P(X = 1) than summing P(X = 2) + P(X = 3) + ... + P(X = n).
Linearity of expectation does not require independence. Students sometimes hesitate to apply E[aX + bY] = aE[X] + bE[Y] when X and Y are dependent, but it always works.
⚠️ Identifying the correct distribution from a word problem is worth 3 points on its own. Read the scenario carefully for the clues: fixed trials + two outcomes = binomial; trials until first success = geometric.
⚠️ The E[X²] integral question requires you to show every step: the definition, substitution, integration, evaluation and final answer. Partial credit is awarded at each stage, so write out each line.
⚠️ Conditional probability questions ("given that X > 1, find P(X = 2)") test whether you can set up the ratio P(A and B) / P(B) correctly.
⚠️ Linearity of expectation is tested as a multi-part question: compute E[X] first, then use it to find E[aX + bY]. Show the formula before plugging in numbers.
You flip a coin 10 times and count heads. This is a ______ distribution. (Binomial.)
True or false: E[X²] is the same as (E[X])². (False.)
For a uniform distribution on [0, 5], the PDF height is ______. (1/5 = 0.2.)
True or false: Linearity of expectation requires X and Y to be independent. (False.)
P(A|B) = P(A and B) / ______. (P(B).)
Q: Betty rolls a fair 8-sided die 5 times. What is the probability that she rolls a "3" more than once?
A: This is binomial with n = 5, p = 1/8 = 0.125. P(X > 1) = 1 - P(X = 0) - P(X = 1) = 1 - (0.875)⁵ - 5(0.125)(0.875)⁴ = 1 - 0.513 - 0.366 = 0.121.
Q: Given that Betty rolls a "3" more than once, what is the probability she rolls it exactly twice?
A: P(X = 2 | X > 1) = P(X = 2) / P(X > 1). P(X = 2) = C(5,2)(0.125)²(0.875)³ = 10(0.016)(0.670) = 0.107. So 0.107 / 0.121 = 0.884.
Q: X is uniform on [0, 2] with PDF f(x) = 1/2. Calculate E[X²].
A: E[X²] = integral from 0 to 2 of x²(1/2) dx = (1/2)[x³/3] from 0 to 2 = (1/2)(8/3) = 4/3 = 1.33.
Q: E[X] = 0.625 and E[Y] = 0.8. What is E[2X - 3Y]?
A: E[2X - 3Y] = 2(0.625) - 3(0.8) = 1.25 - 2.4 = -1.15.
The binomial distribution connects to the normal distribution through the normal approximation to the binomial (when np and n(1 - p) are both large enough). Expected value calculations here prepare you for the variance formula Var(X) = E[X²] - (E[X])², which you will need for hypothesis testing and confidence intervals. Conditional probability is the foundation for Bayes' theorem, which appears later in the course.
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