Difficulty: Intermediate | Prerequisites: Basic probability, expected value, variance
Discrete probability distributions assign probabilities to countable outcomes: the number of successes in a fixed number of trials (binomial), or the number of events in a fixed time window (Poisson). These two distributions appear constantly in STAT 350 and form the backbone of the discrete side of the course. If you can work fluently with their PMFs, means, variances and key properties, you are well placed for the exam.
The binomial distribution counts successes in n independent trials, each with the same probability p. The Poisson distribution counts events occurring at a constant average rate over a fixed interval. Both have clean formulas for their PMFs, means and variances, and exam questions test your ability to compute probabilities, recognise symmetry conditions, and apply conditional probability using these distributions.
Probability mass function (PMF)
A function that gives P(X = x) for each possible value x of a discrete random variable. The PMF must be non-negative everywhere and sum to 1 over all possible values.
Think of it as the complete list of "what can happen and how likely is each outcome."
Binomial distribution, Bin(n, p)
Models the number of successes in n independent Bernoulli trials, each with success probability p.
PMF: P(X = x) = C(n, x) p^x (1 - p)^(n - x), for x = 0, 1, 2, ..., n
Mean: E[X] = np
Variance: Var(X) = np(1 - p)
Support: {0, 1, 2, ..., n}. Z can take any value from 0 to n, regardless of whether p is small or large. A small p makes large values unlikely, but not impossible.
Poisson distribution, Poisson(λ)
Models the number of events occurring in a fixed interval of time (or space), given a constant average rate λ.
PMF: P(X = x) = (λ^x * e^(-λ)) / x!, for x = 0, 1, 2, ...
Mean: E[X] = λ
Variance: Var(X) = λ
Support: {0, 1, 2, 3, ...} (all non-negative integers)
In simple terms, use Poisson when you are counting how many times something happens in a fixed period, and the events occur independently at a steady rate.
Parameter λ (lambda)
For the Poisson distribution, λ is the average rate of occurrence. It can be any positive real number; it does not need to be an integer.
For the exponential distribution (covered separately), λ is the rate parameter as well, but the exponential models the time between events rather than the count of events.
Independence of intervals (Poisson)
Events in non-overlapping time intervals are independent. If X1 counts events in hour 1 and X2 counts events in hour 2, then X1 and X2 are independent Poisson random variables (each with the same λ if the rate is constant).
The binomial distribution is symmetric when p = 0.5.
When p is close to 0 or close to 1, the distribution is skewed (right-skewed when p is near 0, left-skewed when p is near 1).
A common exam trap: stating that the binomial is symmetric when p is close to 0 or 1. That is false.
Z ~ Bin(n, p) can take any integer value from 0 to n, inclusive.
Even if p < 0.5, Z can still exceed n/2. The probability may be small, but it is not zero.
Exam pitfall: a statement claiming Z is "supported only on integers strictly less than n/2" when p < 0.5 is false.
For X ~ Poisson(λ = 1):
P(X = 0) = e^(-1) ≈ 0.3679
P(X = 1) = 1 * e^(-1) / 1! = e^(-1) ≈ 0.3679
P(X = 2) = 1^2 * e^(-1) / 2! = e^(-1) / 2 ≈ 0.1839
P(X > 2) = 1 - P(X = 0) - P(X = 1) - P(X = 2) = 1 - e^(-1) - e^(-1) - e^(-1)/2 = 1 - 5e^(-1)/2 ≈ 0.0803
Note: a common exam answer choice is ≈ 0.2642 for P(X > 2). Check this: 1 - P(X ≤ 2) = 1 - (0.3679 + 0.3679 + 0.1839) = 1 - 0.9197 = 0.0803. So 0.2642 is incorrect for λ = 1.
If X ~ Poisson(λ = 1) counts owls per hour, and hours are independent:
P(1 owl in hour 1 AND 0 owls in hour 2) = P(X1 = 1) P(X2 = 0) = e^(-1) e^(-1) = e^(-2) ≈ 0.1353
This uses the independence of non-overlapping intervals.
If P(X = 5) = P(X = 7), set the PMF expressions equal:
λ^5 e^(-λ) / 5! = λ^7 e^(-λ) / 7!
Cancel e^(-λ) from both sides: λ^5 / 5! = λ^7 / 7!
Simplify: 1 / 5! = λ^2 / 7! → λ^2 = 7! / 5! = 7 * 6 = 42
So λ = √42
Therefore P(X = 0) = e^(-λ) = e^(-√42)
Given two PMFs (one for each scenario) and prior probabilities of each scenario, use the law of total probability and Bayes' theorem.
Worked example from the exam (soccer field problem):
Setup: STAT HS holds the reservation 74% of the time; STAT MS holds it 26% of the time.
STAT HS PMF: P(0) = 0.05, P(1) = 0.20, P(2) = 0.45, P(3) = 0.30
STAT MS PMF: P(0) = 0.40, P(1) = 0.40, P(2) = 0, P(3) = 0.20
(a) P(exactly 3 games | at least 1 game, STAT HS has field)
P(X ≥ 1 | HS) = 1 - P(X = 0 | HS) = 1 - 0.05 = 0.95
P(X = 3 | X ≥ 1, HS) = P(X = 3 | HS) / P(X ≥ 1 | HS) = 0.30 / 0.95 ≈ 0.3158
(b) P(at least 1 game in any given month)
Use total probability across both schools:
P(X ≥ 1) = 1 - P(X = 0)
P(X = 0) = P(X = 0 | HS) P(HS) + P(X = 0 | MS) P(MS)
P(X = 0) = 0.05 0.74 + 0.40 0.26 = 0.037 + 0.104 = 0.141
P(X ≥ 1) = 1 - 0.141 = 0.8590
(c) P(STAT MS holds reservation | at least 1 game)
Bayes' theorem:
P(MS | X ≥ 1) = P(X ≥ 1 | MS) * P(MS) / P(X ≥ 1)
P(X ≥ 1 | MS) = 1 - 0.40 = 0.60
P(MS | X ≥ 1) = (0.60 * 0.26) / 0.8590 = 0.156 / 0.859 ≈ 0.1816
(d) Testing independence of reservation holder and games played
Two events are independent if P(A ∩ B) = P(A) * P(B).
Check: P(X = 0 ∩ HS) = P(X = 0 | HS) P(HS) = 0.05 0.74 = 0.037
P(X = 0) P(HS) = 0.141 0.74 = 0.1043
0.037 ≠ 0.1043, so they are not independent.
The reservation holder and the number of games played are dependent, because the two schools have different PMFs for games.
Binomial PMF: P(X = x) = C(n, x) p^x (1 - p)^(n - x)
Poisson PMF: P(X = x) = λ^x * e^(-λ) / x!
Conditional probability: P(A | B) = P(A ∩ B) / P(B)
Bayes' theorem: P(B | A) = P(A | B) * P(B) / P(A)
Law of total probability: P(A) = Σ P(A | Bi) * P(Bi), summed over all mutually exclusive, exhaustive events Bi
Students often think the Poisson parameter λ must be a positive integer. It can be any positive real number (e.g., λ = 2.7 or λ = √42).
A frequent error is claiming that Bin(n, p) with p < 0.5 cannot produce values above n/2. It can, just with low probability.
Students sometimes forget to use independence of Poisson intervals when computing joint probabilities across separate time windows. Multiply the individual probabilities.
Confusing "at least one" with "exactly one" is common. P(X ≥ 1) = 1 - P(X = 0), not P(X = 1).
⚠️ Conditional probability questions using PMF tables appear as multi-part free response problems. Practice the full Bayes' theorem workflow.
⚠️ The Poisson "set P(X = a) = P(X = b) and solve for λ" problem type is a classic exam question.
⚠️ Know when to use 1 - P(X = 0) for "at least one" computations.
⚠️ Independence tests (P(A ∩ B) vs P(A) * P(B)) are regularly tested in the free response section.
True or false: The binomial distribution is symmetric when p = 0.3.
Fill in the blank: For X ~ Poisson(λ), both the mean and variance equal ______.
True or false: If X ~ Poisson(λ = 1), then P(X > 2) ≈ 0.2642.
True or false: For Bin(n, p) with p < 0.5, the random variable can never take a value greater than n/2.
Fill in the blank: Two events A and B are independent if and only if P(A ∩ B) = ______.
Q: X ~ Poisson(λ) and P(X = 3) = P(X = 5). What is λ?
A: Setting the PMFs equal: λ^3/3! = λ^5/5!. Simplifying gives λ^2 = 5 * 4 = 20, so λ = √20 = 2√5.
Q: A binomial random variable has n = 20 and p = 0.1. Can X take the value 15?
A: Yes. P(X = 15) = C(20,15) 0.1^15 0.9^5, which is extremely small but not zero. The support of Bin(20, 0.1) is {0, 1, 2, ..., 20}.
Q: Events arrive at a Poisson rate of λ = 3 per hour. What is the probability of exactly 0 events in one hour, followed by exactly 2 events in the next hour?
A: By independence of non-overlapping intervals: P(X1 = 0) P(X2 = 2) = e^(-3) (3^2 e^(-3) / 2!) = e^(-3) 9e^(-3)/2 = 9e^(-6)/2 ≈ 0.0112.
Q: Two scenarios lead to different PMFs for the number of goals scored. Scenario A happens 60% of the time with P(0 goals) = 0.1. Scenario B happens 40% of the time with P(0 goals) = 0.5. What is the overall probability of at least 1 goal?
A: P(0 goals) = 0.1 0.6 + 0.5 0.4 = 0.06 + 0.20 = 0.26. So P(at least 1) = 1 - 0.26 = 0.74.
The Poisson distribution connects directly to the exponential distribution: if events arrive at a Poisson rate λ, then the time between consecutive events follows Exp(λ). This relationship is tested in STAT 350.
Conditional probability and Bayes' theorem from this section are used again when working with continuous distributions (e.g., P(X > a | X > b) for the normal distribution).
The binomial distribution connects to the normal distribution through the normal approximation to the binomial (for large n), which may appear later in the course.
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