The Fundamental Counting Principle.
Choices at each stage multiply together.
The tree shows why multiplication is right: every one of the 2 shirts opens up all 3 trousers, so the branches double before they triple.
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- Probability and the Counting Principle
Probability and the Counting Principle
Multiply the choices at each stage. That is how many outcomes there are.
Count the outcomes
A canteen offers 4 main dishes, 3 drinks and 2 desserts. How many different meals of one of each are possible?
Each of the 4 mains can go with any of the 3 drinks, and each of those 12 pairs with either dessert: 4 × 3 × 2 = 24. A and B add instead of multiplying — 4 + 3 + 2 = 9 counts the individual items, not the combinations. Every extra stage multiplies the possibilities.
With and without repetition.
3-digit PIN, repeats allowed
10 × 10 × 10
= 1,000
3 different digits
10 × 9 × 8
= 720
3 runners from 8, in order
8 × 7 × 6
= 336
"Without repetition" simply means each stage has one fewer choice than the last, because something has been used up.
Theoretical probability: reasoning, not trying.
P(event) = favourable outcomes / total outcomes
a bag of 5 red, 3 blue, 4 green
12 marbles
P(blue) = 312 = 14
simplify
P(not blue) = 912 = 34
or 1 − 14
The two add to 1, which is always true of an event and its complement — a useful check on any probability answer.
The standard deck
52 cards in 4 suits of 13: hearts and diamonds are red, clubs and spades are black. Each suit has an ace, 2 to 10, and three face cards (jack, queen, king) — so there are 4 of every rank. P(a red card) = 2652 = 12, and P(a face card) = 1252 = 313.
Experimental probability: what actually happened.
times it occurred / times you tried
a die rolled 60 times
a six came up 14 times
experimental: 1460 ≈ 0.233
what was observed
theoretical: 16 ≈ 0.167
what reasoning predicts
A gap is not evidence of a loaded die
Sixty rolls is a small number. More trials pull the experimental value towards the theoretical one, but they never have to match exactly — and a run of sixes does not make a seventh less likely.
Counting and probability together.
a coin and a 4-section spinner
P(head and a 3)?
total outcomes = 2 × 4 = 8
counting principle
favourable = 1
just H3
P = 18
= (12)(14) ✓
Counting the outcomes and multiplying the separate probabilities give the same answer, because both rest on the same principle.
Two dice, and the 36 outcomes behind them.
Two dice give 6 × 6 = 36 equally likely ordered pairs. Count pairs, not sums.
Two fair dice are rolled. What is the probability that the sum is 9?
pairs with sum 9: (3, 6), (4, 5), (5, 4), (6, 3)
first die, second die
P = 436 = 19
favourable over 36
(4, 5) and (5, 4) are different outcomes
Counting them once gives 236 = 118, half the true value. Think of one red die and one blue die: red 4 with blue 5 is not the same roll as red 5 with blue 4. And the eleven sums 2 to 12 are not equally likely, so 111 is wrong too.
Does order matter? Is the item put back?
Posts with titles: order matters. A plain group: divide out the repeats. Not replaced: the second draw has one fewer.
From 6 learners, two will represent the class (no titles). How many pairs are possible?
6 × 5 = 30
as if order mattered
each pair counted twice: (Ana, Ben) and (Ben, Ana)
the same pair
30 ÷ 2 = 15 pairs
With titles, such as president and secretary, the 30 stays — Ana-then-Ben is a different result from Ben-then-Ana.
A bag holds 3 red and 5 white balls. Two are drawn without putting the first back. What is P(both red)?
P(1st red) = 38
8 balls
P(2nd red) = 27
one red and one ball fewer
P = 38 × 27 = 656 = 328
multiply the stages
A coin is tossed 3 times. What is the probability of at least one head?
P(no heads) = (12)³ = 18
tail, tail, tail
P(at least one head) = 1 − 18 = 78
the complement
"At least one" is easiest through its opposite, "none".
Variations: impossible and certain, a restricted first digit, predicting from an experiment.
The same two ideas, at their edges.
One die is rolled. Find P(7) and P(a number less than 7).
P(7) = 06 = 0
no face shows 7: impossible
P(less than 7) = 66 = 1
every face works: certain
Every probability lies from 0 to 1. An answer outside that range signals a slip.
How many 3-digit numbers have no repeated digit? (The first digit cannot be 0.)
first digit: 9 choices
1 to 9
second: 9 choices
0 is back, one digit used
third: 8 choices
9 × 9 × 8 = 648
Fill the restricted position first. Starting with 10 × 9 × 8 = 720 counts numbers like 045, which are not 3-digit numbers.
A spinner lands on blue 18 times in 60 spins. About how many blues should we expect in 250 spins?
experimental P(blue) = 1860 = 0.3
expected blues ≈ 0.3 × 250 = 75
probability × trials
An experimental probability can be used to predict, but the actual count will vary around 75.
On the exam
Items ask how many outcomes a multi-stage choice has, for a simple probability from a bag, deck or die, for an experimental probability, or for a two-stage probability with dice or draws. The fastest route is to multiply the stages, then ask whether order matters and whether items are put back. For a 10-second check, a probability must lie between 0 and 1, and P(event) + P(not event) = 1.
Now you try
Practice problem
Count the arrangements
How many 3-letter codes can be made from the letters A, B, C, D and E if no letter repeats?
The first letter has 5 choices, the second only 4 (one is used), the third 3: 5 × 4 × 3 = 60. C is 5³, which would be right if repeats were allowed. A multiplies 5 by 3, the letters times the places, instead of counting the choices left at each stage.
Hint: "no letter repeats" means each stage has one fewer option than the one before it.