STAT101 - Introduction to Probability Theory and Statistics
A slide-deck replacement: random variables, expected value, the binomial distribution, and the Poisson distribution — with full worked examples and exam drills.
Calculate and interpret the expected value of a discrete random variable.
Apply the binomial distribution to business decision-making situations.
Compute probabilities for the Poisson distribution and apply it to decision-making situations.
A random variable takes on different numerical values based on chance. There are two flavors you must be able to tell apart:
If you can list the possible values ("0, 1, 2, 3, ...") it's discrete. If the variable is measured on a continuum (any decimal value is possible — length, time, weight, temperature) it's continuous. Counting questions ("how many...") are almost always discrete.
A discrete probability distribution lists every possible value x of the random variable together with its probability P(x). Two requirements must always hold:
Real estate listings show homes with various numbers of bedrooms. Let x = number of bedrooms. The probability of each outcome is based on relative frequency:
| x = number of bedrooms | P(x) |
|---|---|
| 1 | 0.10 |
| 2 | 0.30 |
| 3 | 0.40 |
| 4 | 0.10 |
| 5 | 0.05 |
| 6 | 0.05 |
| Sum | 1.00 |
Every P(x) is between 0 and 1, and they sum to 1.00 — so this is a valid discrete probability distribution.
The expected value (mean) of a discrete probability distribution is the long-run average value you'd see if the underlying experiment were repeated over and over. It is a weighted average of all possible x values, weighted by their probabilities.
Using the bedroom distribution above, compute E(x) column by column:
| x | P(x) | x·P(x) |
|---|---|---|
| 1 | 0.10 | 0.10 |
| 2 | 0.30 | 0.60 |
| 3 | 0.40 | 1.20 |
| 4 | 0.10 | 0.40 |
| 5 | 0.05 | 0.25 |
| 6 | 0.05 | 0.30 |
| E(x) = | 2.85 | |
Answer: The expected number of bedrooms is 2.85 — this is the mean of the probability distribution (note it does not have to be one of the actual possible x values).
You pay $3 to play a game: roll one 6-sided die and receive a payout of $1 times the number that comes up. What is the expected profit (payout minus the $3 cost)?
| x = profit | P(x) | x·P(x) |
|---|---|---|
| −$2 (rolled 1) | 0.167 | −$0.33 |
| −$1 (rolled 2) | 0.167 | −$0.17 |
| $0 (rolled 3) | 0.167 | $0.00 |
| $1 (rolled 4) | 0.167 | $0.17 |
| $2 (rolled 5) | 0.167 | $0.33 |
| $3 (rolled 6) | 0.167 | $0.50 |
| E(x) = | $0.50 | |
Answer: You could expect to win $0.50 on average every time you play — a good deal for you (and a bad deal for whoever's running the game)!
Standard deviation measures the spread, or dispersion, of a random variable's values around its mean.
Continuing the bedroom example (E(x) = 2.85), build all five columns needed for σ by hand:
| x | P(x) | [x − E(x)] | [x − E(x)]² | [x − E(x)]²·P(x) |
|---|---|---|---|---|
| 1 | 0.10 | −1.85 | 3.4225 | 0.342 |
| 2 | 0.30 | −0.85 | 0.7225 | 0.217 |
| 3 | 0.40 | 0.15 | 0.0225 | 0.009 |
| 4 | 0.10 | 1.15 | 1.3225 | 0.132 |
| 5 | 0.05 | 2.15 | 4.6225 | 0.231 |
| 6 | 0.05 | 3.15 | 9.9225 | 0.496 |
| sum (variance) = | 1.427 | |||
Answer: σ = √1.427 = 1.19 bedrooms. This tells us the typical listing's bedroom count deviates from the mean (2.85) by roughly 1.19 bedrooms.
Always build the table left to right in this exact column order: x, P(x), x·P(x) (sum this column to get E(x) first), then [x−E(x)], then square it, then multiply by P(x) and sum that last column for the variance. Doing E(x) first is required — you cannot compute [x−E(x)] until you already know E(x).
Do not forget to take the square root at the end! Σ[x−E(x)]²P(x) gives you the variance (σ²), not the standard deviation. Students very often stop one step early and report the variance as if it were σ.
The binomial distribution gives the probability of x successes in n trials in a process that meets all four of these conditions:
Typical binomial situations: a manufacturer classifies products as defective/acceptable; a firm bidding for a contract gets it or doesn't; survey respondents say "yes I will buy" or "no I will not"; job applicants accept or reject an offer.
where x = number of successes (x = 0, 1, 2, ..., n), n = number of trials, p = probability of success, q = 1 − p = probability of failure.
The term n!/(x!(n−x)!) counts the number of different ways x successes can occur among n trials (order does not matter). Recall:
A bank believes 80% of its customers prefer the bank to close early Friday but open Saturday morning. Three customers are randomly surveyed. Let x = number who prefer Saturday. Find P(x = 3), i.e., the probability all three prefer Saturday.
Here n = 3, p = 0.80, q = 0.20, x = 3.
The full distribution for n = 3 is:
| x = number who prefer Saturday | P(x) |
|---|---|
| 0 | 0.008 |
| 1 | 0.096 |
| 2 | 0.384 |
| 3 | 0.512 |
| Sum | 1.000 |
Answer: P(x = 3) = 0.512, i.e. a 51.2% chance all three surveyed customers prefer the Saturday option.
Now suppose the bank interviews n = 8 customers (still p = 0.80). What is the probability that at least 6 of the 8 prefer the Saturday option?
"At least 6" means x = 6, 7, or 8. Using the binomial formula for each x (or the binomial table for n = 8, p = 0.80):
| x | P(x) |
|---|---|
| 0 | 0.000 |
| 1 | 0.000 |
| 2 | 0.001 |
| 3 | 0.009 |
| 4 | 0.046 |
| 5 | 0.147 |
| 6 | 0.294 |
| 7 | 0.336 |
| 8 | 0.168 |
| Sum | 1.000 |
Answer: There is a 79.8% chance that at least 6 of the 8 customers surveyed prefer the Saturday option.
Mean and standard deviation for this n = 8 binomial: μ = np = 8(0.80) = 6.40, σ = √(npq) = √(8·0.80·0.20) = √1.28 = 1.13.
An airline knows that the probability a first-class ticket holder is a no-show is p = 0.08. The airline sold n = 10 first-class tickets. What is the probability that exactly 2 people are no-shows?
From the cumulative binomial table (n = 10, p = 0.08): P(x ≤ 2) = 0.9599 and P(x ≤ 1) = 0.8121.
Answer: P(exactly 2 no-shows) ≈ 0.1478, or about 14.8%.
A contract for 3,000 components uses this acceptance sampling plan: inspect n = 20 components; if x ≤ 1 defect, accept the shipment; if x > 1, reject it.
If the shipment is actually good (true defect rate p = 0.05): from the cumulative binomial table (n = 20, p = 0.05), P(x ≤ 1) = 0.7358 → 73.58% chance of correctly accepting a good shipment.
If the shipment is actually bad (true defect rate p = 0.10): from the table (n = 20, p = 0.10), P(x ≤ 1) = 0.3917 → 39.17% chance of incorrectly accepting a bad shipment.
Answer: This sampling plan does a mediocre job — it only accepts good shipments 73.6% of the time (we'd like this closer to 1) while still accepting bad shipments nearly 40% of the time (we'd like this closer to 0). A better plan would use a larger n or a stricter acceptance number.
These shortcut formulas match the long way (Σx·P(x) and Σ[x−E(x)]²P(x)) — for the n = 8 bank example, μ = np = 8(0.80) = 6.40 and σ² = npq = 8(0.80)(0.20) = 1.28, so σ = √1.28 ≈ 1.13, consistent with computing E(x) and the variance column by column from the full probability table.
The Poisson distribution describes a process that extends over space, time, or any well-defined segment/interval, in which the outcomes of interest occur at random and the number of outcomes in a given interval is counted. It's used when the total number of possible outcomes cannot be determined (unlike the binomial, there's no fixed "n" trials).
Characteristics:
Classic Poisson examples: number of patients arriving at an ER per hour; number of tech-support calls received in a 30-minute period; number of cars served at a gas station in 24 hours; number of defects per square meter of leather.
where t = number of segments of interest, x = number of successes in t segments, λ = expected number of successes in one segment, e = base of the natural logarithm (≈ 2.71828). Mean/expected value = λt.
At a Major League Baseball park, 5 customers arrive at a concession stand on average in a 10-minute period. What is the probability that in a 10-minute period, exactly 3 people will arrive?
Segment size = 10 minutes, λ = 5 (per 10-min segment), t = 1 segment, so λt = 5.
Answer: P(exactly 3 arrivals in 10 minutes) ≈ 0.1404, or about 14.0%.
Same concession stand (λ = 5 per 10 minutes). Find the probability of 3 or fewer customers arriving in a 20-minute period.
Now t = 2 segments (since 20 minutes = two 10-minute segments), so λt = 5 × 2 = 10.
Using the cumulative Poisson table with λt = 10.00 and x = 3: P(x ≤ 3) = 0.0103.
Answer: There is only about a 1.03% chance of 3 or fewer arrivals in 20 minutes — makes sense, since we'd expect 10 arrivals on average in that longer window, so getting only 3 or fewer would be unusually low.
The number of arrivals to the checkout section of a store per hour follows a Poisson distribution with mean λ = 16 (t = 1 hour). What is the probability of exactly 12 customers arriving in one hour?
Answer: P(x = 12) = 0.0661, about 6.6%. Notice the distribution is centered near the mean of 16 and looks roughly bell-shaped (Poisson distributions become more symmetric as λt grows larger).
| Clue in the word problem | Use... |
|---|---|
| Fixed number of trials n, each with a clear success/failure outcome ("out of 20 shipped parts", "3 customers surveyed") | Binomial |
| A rate of occurrence over a continuous span of time, area, or volume, with no natural "n" ("per hour", "per square meter", "in a 10-minute period") | Poisson |
| You're told "p = probability of success" and a specific sample size n | Binomial |
| You're told "on average λ events happen per interval" with no upper bound on how many could occur | Poisson |
1. Which of the following is a requirement of a valid discrete probability distribution?
2. A discrete random variable x has E(x) = 40. What does this value represent?
3. Which of the following is NOT one of the four required conditions for a binomial distribution?
4. In the binomial formula P(x) = [n!/(x!(n−x)!)]·pˣ·q^(n−x), what does q represent?
5. A binomial experiment has n = 12 and p = 0.30. What is the mean μ?
6. Which scenario is best modeled by a Poisson distribution rather than a binomial distribution?
7. In the Poisson formula P(x) = (λt)ˣ·e^(−λt)/x!, what does λ represent?
8. A call center averages 4 calls every 10 minutes (Poisson). If asked for the probability of receiving exactly 6 calls in a 20-minute period, what value of λt should be used in the formula?
9. If you are given P(x ≤ 4) = 0.72 from a cumulative binomial table, how would you find P(x ≥ 5)?
10. A discrete random variable has the distribution: P(0) = 0.20, P(1) = 0.50, P(2) = 0.30. Compute E(x) and the standard deviation σ.
Answer:
E(x) = 0(0.20) + 1(0.50) + 2(0.30) = 0 + 0.50 + 0.60 = 1.10
[x−E(x)]² · P(x): for x=0: (0−1.1)² = 1.21, ×0.20 = 0.242. For x=1: (1−1.1)² = 0.01, ×0.50 = 0.005. For x=2: (2−1.1)² = 0.81, ×0.30 = 0.243.
Sum = 0.242 + 0.005 + 0.243 = 0.490 (this is the variance)
σ = √0.490 = 0.70
11. A production line has a 10% defect rate (p = 0.10). A sample of n = 6 items is inspected. Use the binomial formula to find the probability that exactly 2 items are defective.
Answer:
n = 6, x = 2, p = 0.10, q = 0.90
P(2) = [6!/(2!·4!)] · (0.10)² · (0.90)⁴ = 15 · 0.01 · 0.6561
P(2) = 15 × 0.006561 = 0.0984
There is about a 9.84% chance exactly 2 of the 6 sampled items are defective.
12. A hospital ER receives an average of 3 patients per 15-minute period (Poisson). Find the probability that exactly 2 patients arrive in a 15-minute period.
Answer:
λ = 3 per 15-minute segment, t = 1, so λt = 3.
P(x=2) = 3² · e⁻³ / 2! = 9 · 0.049787 / 2 = 0.4481 / 2
P(x=2) = 0.2240
There is about a 22.4% chance exactly 2 patients arrive in that 15-minute period.
| Term | Definition |
|---|---|
| Random variable | A variable that takes on different numerical values based on chance. |
| Discrete probability distribution | A table/function listing every possible value of a discrete random variable along with its probability, where 0 ≤ P(x) ≤ 1 and Σ P(x) = 1. |
| Expected value | The mean, μ = E(x) = Σ x·P(x), of a discrete probability distribution — the long-run average outcome. |
| Binomial distribution | The distribution of the number of successes x in a fixed number n of independent, identical trials, each with constant success probability p. |
| Poisson distribution | The distribution of the number of rare, random outcomes occurring in a fixed interval of time or space, with mean λt. |
| Trial | A single repetition of the experiment/process in a binomial setting (e.g., inspecting one item, surveying one customer). |
| Success / Failure | The two possible outcomes of a single binomial trial; p = P(success), q = 1 − p = P(failure). |