PSYB06H3 Chap.4 Counting Outcomes and Binomial Probability
Counting Outcomes and Binomial Probability
Two questions pick the counting rule
Counting problems begin by asking whether order matters and whether an item may be chosen again. Ordered selections are permutations and unordered ones are combinations, and each comes in a version with repeats and a version without. Factorials describe how many ways a set can be arranged: each position filled leaves one fewer option for the next.
Writing those two answers beside the question makes the right formula on the sheet obvious rather than a guess. A four-digit keypad code, for example, keeps order and allows repeated digits, whereas a committee drawn from a class keeps neither order nor repeats.
Recognising those two cases quickly is the habit tested by short counting questions on quizzes.
Binomial probability is built from simpler rules
A binomial question asks for a number of successes across repeated independent events, such as correct guesses on a quiz or heads across coin flips. One specific sequence is priced with the multiplicative rule, P for each success and Q for each failure.
Every sequence with the same number of successes has that same chance, so the combination count multiplies it. The formula p equals C times P to the X times Q to the N minus X is a compact record of those steps.
Exactly, at least and the exhaustive shortcut
The wording of a question changes the arithmetic. Exactly four heads is one bar of the distribution; at least four heads collects every bar from four upward.
When the wanted range holds many outcomes, add the short list outside it and subtract from one, remembering that zero belongs to that list.
The probability of at least one occurrence across N events reduces to one minus the chance of none, which is Q raised to N.
Replacement changes the model
The binomial model needs the chance of success to stay the same on every event, which holds when each draw is returned before the next.
Drawing without replacement shrinks the pool, so each outcome updates the next probability and the calculation becomes a product of changing fractions. With many events, the binomial distribution has mean NP and variance NPQ, which later lets a z score approximate a ranged probability.
The expected number of successes, NP, also gives a quick sense check: a computed distribution whose tallest bar sits far from NP signals an arithmetic slip somewhere in the powers or the counts.
What this chapter covers
- 01
Permutations With and Without Repeats
- 02
Combinations and Factorials
- 03
Binomial Probability Formula
- 04
Exactly Versus At Least
- 05
Exhaustive Complement Shortcut
- 06
Sampling With and Without Replacement
Worked example · free
Exactly two and at least two of six trials
- 1Count the orders: C = 6! / (2! × 4!) = 15.
- 1Price one sequence: 0.20² × 0.80⁴ = 0.04 × 0.4096 = 0.016384, so p(2/6) = 15 × 0.016384 = 0.2458.
- 1Find the outcomes outside the range: p(0/6) = 0.80⁶ = 0.2621 and p(1/6) = 6 × 0.20 × 0.80⁵ = 0.3932.
- 1Apply the complement: p(at least 2) = 1 − 0.2621 − 0.3932 = 0.3446.
Key terms
- Permutation
- A count of arrangements in which the order of the selected items matters.
- Combination
- A count of selections in which order does not matter, so rearrangements of the same items are counted once.
- Factorial
- The product of a whole number and every positive whole number below it, giving the ways to order a full set.
- Binomial Distribution
- The probability distribution of the number of successes across N independent events with a fixed chance of success.
- Exhaustive Outcomes
- A set of outcomes that covers every possibility, so their probabilities sum to one.
- Sequential Probability
- Probability for draws without replacement, where each outcome changes the chances for the next draw.
Counting Outcomes and Binomial Probability FAQ
How do I decide between a permutation and a combination?
Ask whether swapping two selected items creates a different outcome. If assigning the same people to different named roles counts as a new result, order matters and a permutation applies. If only the membership of the group matters, use a combination, which divides out the orderings of the chosen items.
Why does the binomial formula multiply by the number of combinations?
Each specific sequence with X successes has the same chance, P to the X times Q to the N minus X. Those sequences are mutually exclusive, so their probabilities add, and adding the same value C times is the same as multiplying by C. Leaving out that count gives the chance of only one ordering.
When is the one minus approach faster than adding outcomes?
It is faster whenever the outcomes you want outnumber the ones you do not. For at least one success, the only excluded outcome is zero successes, so one minus Q to the N finishes the problem. For a range like more than five out of fifty, add zero to five and subtract the total from one.
Why does drawing without replacement break the binomial model?
The binomial formula assumes every event has the same chance of success. Removing an item from a small pool changes that chance for the next draw, so the events are no longer identical. Multiply the changing fractions draw by draw instead, and multiply by the number of qualifying orders when more than one sequence fits.
Exam move
Write the two deciding questions, order and repeats, at the top of every counting problem before choosing a formula. Practise the binomial in three written lines: the count of orders, the price of one sequence and their product. Then rework the same scenario as an at-least question using the complement and check that the answer is larger.
Finish with one problem drawn without replacement so the contrast between fixed and changing probabilities stays clear.
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