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The product rule, what an outcome is, and why counting the complement is usually shorter.
Paper packet. Every task here also exists on screen, where it is checked automatically; answers written on paper are not assessed by Nydus. When you are back at a device, enter your answers there.
You will say what one outcome of an experiment is, check whether outcomes described that way are equally likely, and count both the sample space and an event with the product rule. You will also recognise when counting the complement is the shorter route, and when re-describing an outcome has quietly destroyed the equally likely model.
When $\Omega$ is finite and the outcomes are equally likely, $P(A) = |A| / |\Omega|$. That is the only bridge between the axioms and a number, and it turns every question of this kind into two counting questions: how many outcomes are there, and how many of them are in the event.
An outcome is one complete result of performing the experiment once, and the sample space is the set of them. A model is equally likely when every outcome carries the same probability. The product rule multiplies the number of possibilities at each stage of a choice. Complementary counting counts what you do not want and subtracts it from the whole.
An equally likely model says the outcomes of $\Omega$ all carry the same probability. It is a modelling assumption, never a consequence of being able to list them, and it has to be argued for from the apparatus: a fair coin, a well-shuffled deck, a point chosen with no preference.
Under that assumption the whole subject is the product rule: if a choice is made in stages, and stage $i$ has $n_i$ possibilities whatever the earlier stages did, the number of outcomes is $n_1 n_2 \cdots n_k$. Two dice give $6 \times 6 = 36$; a four-digit code gives $10^4$; drawing two cards in order from a deck gives $52 \times 51$, because the second stage has one fewer possibility whatever the first was.
The second move is to count the complement. At least one is almost always the harder side: the direct count splits into overlapping cases, while none is a single product. $P(\text{at least one six in five rolls}) = 1 - (5/6)^5$ is one line; the direct count is five.
Another way: story
A restaurant offers three starters, four mains and two puddings. There are $3 \times 4 \times 2 = 24$ meals — and that is the product rule in its whole generality. Everything harder in this unit is the same picture with the later stages depending on the earlier ones.
Another way: steps
The $36$ ordered pairs are equally likely. The eleven totals are not, because a total collects different numbers of pairs.
| Total | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 |
|---|---|---|---|---|---|---|---|---|---|---|---|
| Pairs | 1 | 2 | 3 | 4 | 5 | 6 | 5 | 4 | 3 | 2 | 1 |
The counts add to $36$, as they must. The lesson is general: an equally likely model belongs to a particular description of an outcome, and re-describing the outcome — a total instead of a pair — usually destroys it. Most wrong answers in this unit are this mistake, not an arithmetic slip.
Counting unordered when the model is ordered. If $\Omega$ is the $36$ ordered pairs, the event one die shows 2 and the other shows 5 holds two of them, not one. Mixing an ordered space with an unordered count is the single commonest source of a factor of two.
Assuming the outcomes you happened to list are equally likely. Win, draw or lose is a fine sample space and is not one third each.
*Counting at least one directly.* The cases overlap, so adding them double-counts. Count the complement.
Using the product rule when a stage's count depends on the earlier stages. It needs stage $i$ to have $n_i$ possibilities whatever happened before. Drawing without replacement satisfies that ($51$ either way); drawing until a condition holds does not.
One outcome is an ordered string of four digits, and each position is filled from ten digits whatever the others hold.
Say what an outcome is first.
$|\Omega| = 10^4 = 10000$, and the codes with no repeated digit number $10 \times 9 \times 8 \times 7 = 5040$.
Product rule both times.
So a code chosen at random repeats a digit with probability $1 - 5040/10000 = 0.496$.
The complement was the shorter count.
$|\Omega| = 6^4 = 1296$ ordered quadruples.
Each roll is a free stage.
No six is $5^4 = 625$ of them.
One product, no cases.
So the answer is $1 - 625/1296 = 671/1296$, a little over a half.
The direct count would have been four overlapping cases.
An outcome is an ordered triple of heads and tails, so $|\Omega| = 2^3 = 8$.
Product rule.
No head is one outcome: tails, tails, tails.
The complement is a single outcome.
So the probability is $1 - 1/8 = 7/8$.
Two fair six-sided dice are rolled and the sample space is the $36$ ordered pairs. How many of those pairs give each of these totals?
| Number of ordered pairs | |
|---|---|
| Total of $2$ | |
| Total of $4$ | |
| Total of $7$ |
Which counting model answers this: how many five-card hands can be dealt from a deck of fifty-two?
Two fair six-sided dice are rolled. In how many of the $36$ ordered pairs is the first die strictly smaller than the second, with both dice showing at most $4$?
Answer:
Two fair dice with $7$ faces each are rolled and the faces are read in order. Complete the two counts.
The sample space has n ordered outcomes, and m of them are doubles.
Here is a short argument in four sentences. Mark the two sentences that help themselves to an equally likely model without earning it.
This task has no paper form; do it on a device.
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
Two fair six-sided dice are rolled. Put these four events in order, fewest ordered pairs first.
Number the steps in order (write the number in the box):
You can count a sample space and an event with the product rule and switch to the complement when that is shorter. Say in your own words why the thirty-six ordered pairs of two dice are equally likely while the eleven totals are not.
9. Your turn: three fair coins are tossed; find the probability of at least one head, step 3