MATH 321 – Mathematical Statistics I
October 17, 2026
Example: I offer a lottery game for $5 a play. I spin a spinner with 1-10 on it, evenly spread out. If it lands on 1-9, I keep your money. If it lands on 10, you win $100. Let \(X\) be the payout you receive from this game.
What is the probability mass function of the random variable \(X\)?
What is the expected value of the random variable \(X\), \(E(X)\)?
There are many situations like this, where we have a random experiment with two outcomes, a “success” and a “failure”, with a certain probability of success. Is there a way we can take advantage of the similarities?
These kinds of random variables can be described using a Bernoulli distribution.
A Bernoulli random variable is a random variable with two outcomes, a “success”, 1; or a “failure”, 0, with one parameter: success probability \(p\).
We write this as: \[ X \sim \text{Bernoulli}(p) \]
What does the probability mass function of a \(\text{Bernoulli}(p)\) look like?
The probability mass function of the Bernoulli distribution with success probability \(p\) is: \[ p_X(x) = P(X = x) = \begin{cases} p & \text{for } x = 1 \\ 1 - p & \text{for } x = 0 \end{cases} \]
The expected value of a \(\text{Bernoulli}(p)\) distribution is: \[ E(X) = p. \]
The variance of a \(\text{Bernoulli}(p)\) distribution is: \[ V(X) = p(1-p). \]
Can you prove these?
Example: You flip a biased coin, landing on heads 65% of the time. If the coin lands on heads, you make $5, and if it lands on tails, you get nothing. Let \(X\) be the payout of the game.
Suppose instead of flipping a single coin, we flip it 10 times, and record the number of heads, and call that variable \(X\). How is \(X\) distributed?
So instead of being a single Bernoulli experiment, we are performing 10 independent Bernoulli experiments, and recording the number of “successes”.
This is the idea behind the binomial distribution.
The binomial distribution is defined as the sum of \(n\) independent and identically distributed Bernoulli random variables with success probability \(p\).
In formal terms:
\[ \begin{gather*} X_i \overset{i.i.d.}{\sim} \text{Bernoulli}(p) \\[0.2em] X = \sum_{i=1}^n X_i \\[0.2em] X \sim \text{Binomial}(n, p) \end{gather*} \]
The binomial distribution has \(2\) parameters: the number of trials \(n\) and the success probability \(p\).
The probability mass function of a binomially distributed random variable \(X\) is: \[ p_X(x) = P(X = x) = \binom{n}{x} p^x (1-p)^{n-x}, \] for \(x \in \{ 0, 1, 2, \dots, n \}\).
There are times when the probability mass function isn’t the most useful way to frame a problem.
The cumulative distribution function (CDF) is defined as the probability that the random variable \(X\) is less than or equal to \(x\).
For a discrete random variable: \[ F_X(x) = P(X \le x) = \sum_{k \le x} p_X(k). \]
Example:
The probability that a patient recovers from a stomach disease is \(.8\). Suppose \(20\) people are known to have contracted this disease. What is the probability that:
The expected value of a \(\text{Binomial}(n, p)\) distribution is: \[ E(X) = np. \]
The variance of a \(\text{Binomial}(n, p)\) distribution is: \[ V(X) = np(1-p). \]
Can you prove these?
Example:
An oil exploration firm is formed with enough capital to finance ten explorations. The probability of a particular exploration being successful is \(.1\). Assume the explorations are independent. Find the mean and variance of the number of successful explorations.
Suppose the firm has a fixed cost of $20,000 in preparing equipment prior to doing its first exploration. If each successful exploration costs $30,000 and each unsuccessful exploration costs $15,000, find the expected total cost to the firm for its ten explorations.
Let’s consider a new game: We are going to flip the coin until we get a tails. We will pay out $1 for each coin flip it takes until we get a tails.
Let \(X\) be the payout of this game.
The geometric distribution describes the count of the number of trials until the first success, if each trial is independent and has the same probability of success \(p\).
The geometric distribution has only one parameter, the success probability \(p\).
The probability mass function of a random variable \(X\) with a geometric distribution is: \[ p_X(x) = P(X = x) = (1-p)^{x-1} p, \]
for \(x \in \{ 1, 2, 3, \dots \}\).
Example:
Consider the game from earlier. Suppose the coin lands on tails \(70\%\) of the time. What is the probability that:
The expected value of a \(\text{Geometric}(p)\) distribution is: \[ E(X) = \frac{1}{p}. \]
The variance of a \(\text{Geometric}(p)\) distribution is: \[ V(X) = \frac{1-p}{p^2}. \]
Can you prove these? (Consider a recursive definition. The textbook also uses an argument based on the derivative of the sum of a geometric series.)
Example:
Of a population of consumers, \(60\%\) are reputed to prefer a particular brand, \(A\), of toothpaste. If a group of randomly selected consumers is interviewed, what is the mean and standard deviation of the number of consumers that must be interviewed until someone who doesn’t prefer brand \(A\) is found?
We can further generalize the idea behind the geometric distribution. What if instead of waiting until the first success, we wait until the second success, or the third success, or the \(r^{th}\) success?
The negative binomial distribution describes the count of the number of trials until the \(r^{th}\) success, if each trial is independent and has the same probability of success \(p\).
The negative binomial distribution has two parameters, the number of successes \(r\) and success probability \(p\).
The probability mass function of a random variable \(X\) with a negative binomial distribution is: \[ p_X(x) = P(X = x) = \binom{x - 1}{r - 1} (1-p)^{x-r} p^r, \]
for \(x \in \{ r, r+1, r+2, \dots \}\).
Note that \(X \sim \text{NB}(1, p)\) is the same as \(X \sim \text{Geometric}(p)\).
The expected value of a \(\text{NB}(r, p)\) distribution is: \[ E(X) = \frac{r}{p}. \]
The variance of a \(\text{NB}(r, p)\) distribution is: \[ V(X) = \frac{r(1-p)}{p^2}. \]
A geological study indicates that an exploratory oil well should strike oil with probability \(.2\).
Consider an urn with 8 red balls and 7 black balls in it. If I draw 5 balls from the urn, without replacement, what is the probability that the majority of the balls we pulled out are black?
Let \(Y\) be the number of black balls drawn from the urn in the five balls drawn.
The hypergeometric distribution describes the probability of having \(y\) successes in \(n\) draws, without replacement, from a finite population of size \(N\) that contains exactly \(r\) successes.
The hypergeometric distribution has three parameters, the total number of objects in the population, \(N\), the number of “successful” objects in the population, \(r\), and the number of objects drawn, \(n\).
The probability mass function of a random variable \(Y\) with a hypergeometric distribution is: \[ p_Y(y) = P(Y = y) = \frac{\binom{r}{y} \binom{N-r}{n-y}}{\binom{N}{n}}, \]
where \(y\) is an integer \(\{0,\, 1,\, 2,\, \dots,\, n\}\), subject to the restrictions \(y \le r\) and \(n − y \le N − r\).
The expected value of a \(\text{Hypergeometric}(N, r, n)\) distribution is: \[ E(Y) = \frac{nr}{N}. \]
The variance of a \(\text{Hypergeometric}(N, r, n)\) distribution is: \[ V(Y) = n \left( \frac{r}{N} \right) \left( \frac{N-r}{N} \right) \left( \frac{N-n}{N-1} \right). \]
The proof requires some algebra and facts about the binomial distribution, so I won’t show it here, but it’s not too hard to understand.
Example:
An urn contains ten marbles, of which five are green, two are blue, and three are red. Three marbles are to be drawn from the urn, one at a time without replacement.
The manager at McDonald’s called me to ask for some help modelling his sales. In particular, he wants to predict the number of customers that will come through the drive-thru in an hour, so he can properly staff the window.
Let \(X\) be the number of customers that come through the drive-thru in an hour.
The Poisson distribution expresses the probability of a given number of events occurring in a fixed interval of time or space if these events occur with a known constant mean rate and independently of the time since the last event.
The Poisson distribution has one parameter, the average rate of events, \(\lambda\).
The probability mass function of a random variable \(X\) with a Poisson distribution is: \[ p_X(x) = P(X = x) = \frac{\lambda^x \, e^{-\lambda}}{x!}, \]
where \(x\) is a non-negative integer and \(\lambda > 0\).
The expected value of a \(\text{Poisson}(\lambda)\) distribution is: \[ E(X) = \lambda. \]
The variance of a \(\text{Poisson}(\lambda)\) distribution is: \[ V(X) = \lambda. \]
How might you prove this?
Example:
A certain type of tree has seedlings randomly dispersed in a large area, with the mean density of seedlings being approximately five per square yard.