Theory
From one toss to the whole pattern
Last lesson answered single questions: P(two heads) = 1/4.
New kind of question: a student blind-guesses all 4 options-of-4 MCQs in a quiz. How likely are 0 correct? Exactly 1? Exactly 2? All 4?
Not one probability: a complete table of every possible outcome with its probability. That table has a name, a probability distribution, and two celebrity distributions run this subject: one for counting, one for measuring.
Theory
A menu of chances
A distribution is a menu: every dish (possible value) listed with its price (probability), and the prices always total exactly 1.
Two menu styles exist:
- Discrete (the binomial): finitely many dishes: 0, 1, 2, 3, 4 correct guesses: each with its own price tag.
- Continuous (the normal): infinitely fine dishes (height 167.4938... cm): single dishes are free (P = 0); you pay for ranges: the area under a curve.
Theory
The binomial: counting successes
Use the binomial when ALL four conditions hold:
- fixed number of trials n,
- each trial has two outcomes (success/failure),
- success probability p is the same every trial,
- trials are independent.
Then: P(X = k) = C(n, k) · pᵏ · (1-p)ⁿ⁻ᵏ
C(n, k) = n! / (k!(n-k)!) counts the arrangements (which k of the n trials succeeded). Mean = np, variance = np(1-p): worth memorising as a pair.
Theory
Worked: the blind guesser
n = 4 questions, p = 1/4 (four options each). P(exactly 2 correct)?
1. C(4, 2) = 6 (six ways to choose WHICH two are right).
2. p² = (0.25)² = 0.0625.
3. (1-p)² = (0.75)² = 0.5625.
4. P = 6 × 0.0625 × 0.5625 ≈ 0.211.
About a 21% chance. Mean correct = np = 4 × 0.25 = 1: blind guessing a 4-question quiz typically earns one mark. The full menu (k = 0...4) sums to exactly 1: the distribution's self-check.
Theory
The normal: measuring nature
Plot the CampusPulse heights as a histogram: a symmetric hump: most students near the middle, few very short or very tall. Smooth that hump and you get the normal distribution: the bell curve.
Its identity card:
- Continuous, symmetric, bell-shaped.
- Two parameters: mean μ (where the peak sits) and SD σ (how wide the bell spreads).
- Mean = median = mode, all at the centre.
- Total area under the curve = 1; probability = area over a range.
Heights, measurement errors, and large-class exam marks all sit approximately normal: nature's favourite shape.
Quiz
Height is modelled as normal. A student computes P(height = EXACTLY 170.000 cm) and gets a positive number. What is wrong?
- For a continuous variable, any exact single value has probability 0: only RANGES (area under the curve) carry probability
- Nothing: exact values always have positive probability
- 170 is impossible because it is not the mean
- The normal distribution only handles marks, not heights
Show the answer
For a continuous variable, any exact single value has probability 0: only RANGES (area under the curve) carry probability
A continuous scale has infinitely many values, so the probability mass on any single exact point is zero: you ask P(169.5 ≤ height ≤ 170.5) instead, an area under the bell. This discrete-vs-continuous distinction is THE conceptual divide between the binomial (bars with real heights at each k) and the normal (a curve where only areas mean anything), and exams test it in exactly this form.
Think first
Pick the distribution, twice
Two situations: (1) 10 independent phone calls to parents, each answered with probability 0.6: the NUMBER answered; (2) the exact WEIGHT of rice bags filled by a machine set to 5 kg. Before tapping: which distribution models each, and name the giveaway.
Show the answer
(1) Binomial with n = 10, p = 0.6: fixed trials, two outcomes, constant p, independent: all four conditions checked, and the variable COUNTS successes.
(2) Normal: weight is a continuous MEASUREMENT clustering symmetrically around the 5 kg setting with small random errors either side.
The reflex: counting successes in repeated yes/no trials → binomial; measuring a continuous quantity around a typical value → normal.
Watch out
Condition-checking is the exam
Binomial misuse: drawing 4 cards WITHOUT replacement is not binomial: p changes each draw (independence broken). Say which condition fails.
Formula slips: forgetting C(n, k) (there are SIX ways to get 2-of-4, not one), or swapping the exponents on p and (1-p).
Normal misuse: quoting a positive probability for an exact value: ranges only.
Theory
The bridge between the two
Draw binomial bar-menus for n = 4, then 20, then 100: the bars melt into a smooth, symmetric bell. For large n, the binomial is approximated by the normal: the first hint of a deep pattern (averages of many small chances turn normal) that becomes the Central Limit Theorem two lessons ahead. The bell curve lesson next gives you its working ruler: the 68-95-99.7 rule.
Summary
Key takeaways
- A distribution lists every possible value with its probability; the probabilities total 1.
- Binomial: counts successes in n independent two-outcome trials with constant p.
- P(X = k) = C(n,k) pᵏ (1-p)ⁿ⁻ᵏ; mean np, variance np(1-p).
- Normal: continuous symmetric bell; parameters μ (centre) and σ (width); mean = median = mode.
- Continuous variables: exact values have P = 0; probability lives in areas over ranges.
- Large-n binomials look normal: the CLT preview.
- Memory hook: a menu of chances: priced dishes vs priced ranges.