Short on time? This page is the recap. For explanations and worked examples, use the full study guide.
Recap
- A random variable (capital X) gives a numerical value to each outcome of a random process. Small x is one particular value.
- Discrete random variables have values you can list (0, 1, 2, …). Continuous ones take any value in an interval (Topic 2.11).
- A probability distribution lists every possible value of X with its probability.
- It can be shown as a table, a probability histogram or a function.
- The cumulative distribution gives P(X ≤ x) for each x.
- The exact distribution comes from probability rules; a simulation estimates it with relative frequencies.
Key relationships
| Idea | Rule |
|---|---|
| Valid distribution | 0 ≤ P(X = x) ≤ 1 for every x, and ΣP(X = x) = 1 |
| Missing probability | P(X = one value) = 1 − (sum of all the others) |
| Unknown constant in a function | set the sum of all probabilities equal to 1 and solve |
| Cumulative probability | P(X ≤ x) = sum of P(X = t) for all values t ≤ x |
| Single value from cumulative | P(X = x) = P(X ≤ x) − P(X ≤ previous value) |
| “At least” | P(X ≥ a) = 1 − P(X ≤ value just below a) |
| “Between” | P(a < X ≤ b) = P(X ≤ b) − P(X ≤ a) |
| Simulation estimate | estimated P(X = x) = (trials with value x) ÷ (total trials) |
Assumptions
- Building a distribution with rules uses the rules’ assumptions, often independence. State them.
- A simulation must imitate the real process: correct probabilities, and independence between parts when the real process has it.
- Each trial of a simulation gives one value of X.
Mistakes to avoid
- Checking only the sum. Also check that no probability is negative or above 1.
- Leaving out a value, often 0.
- One row per outcome instead of one per value of X.
- Assuming the values are equally likely.
- Mixing up < and ≤. “Fewer than 3” excludes 3.
- Reading the cumulative table as P(X = x).
- Calling a simulation estimate exact.
Quick self-check
- X takes values 0, 1, 2, 3 with probabilities 0.2, 0.35, k, 0.15. Find k. (k = 1 − 0.70 = 0.3)
- A cumulative table gives P(X ≤ 1) = 0.3, P(X ≤ 2) = 0.65, P(X ≤ 3) = 0.9, P(X ≤ 4) = 1. Find P(X = 2) and P(X > 2). (0.65 − 0.3 = 0.35; 1 − 0.65 = 0.35)
- Is P(X = x) = x ÷ 6 for x = 1, 2, 3 a valid distribution? (Yes: 1/6 + 2/6 + 3/6 = 1, and each value is between 0 and 1)
Next: practice questions.