Equally likely outcomes and complementary events

Theoretical probability when outcomes are equally likely, the two-dice sample space, complementary events, and the shortcut P(not A) = 1 − P(A) for "at least one" questions.

When every outcome in the sample space has the same chance, probability becomes a counting job. Fair coins, fair dice, a well-shuffled set of name cards, a ball drawn blind from a barrel: these are the situations where you can work out a probability without doing a single experiment. The skill is making sure the outcomes you count really are equally likely.

Probability by counting

Equally likely outcomes

$$P(E) = \frac{\text{number of outcomes in the event}}{\text{total number of outcomes}} = \frac{n(E)}{n(S)}$$

Why does this work? If there are 20 equally likely outcomes, each gets an equal share of the certainty, so each has probability $\frac{1}{20}$. An event made of 5 of them gets 5 shares: $\frac{5}{20}$. The formula is just adding up equal shares.

Example 1●

The letters of PARRAMATTA are written on 10 cards and one card is picked at random. Find the probability that it shows (a) an A, (b) a T, (c) a vowel.

$n(S) = 10$

There are 10 cards, all equally likely to be picked, so the total is 10 even though some letters repeat.

$P(A) = \frac{4}{10} = \frac{2}{5}$

Count the A cards: pArrAmAttA has four.

$P(T) = \frac{2}{10} = \frac{1}{5}$

There are two T cards.

$P(\text{vowel}) = \frac{4}{10} = \frac{2}{5}$

The only vowel in the word is A, so the vowel event is the same as the A event.

Two dice: the 36-outcome grid

Roll a red die and a blue die. Each die has 6 faces, so there are $6 \times 6 = 36$ equally likely ordered outcomes. A grid shows them all at once. Totals are not equally likely: a total of 7 can happen six ways, but a total of 12 only one way (6 and 6). That is why you count cells, not totals.

InteractiveShade the two-dice grid

Try this

Choose an event and watch which cells light up. The readouts give the probability as a fraction of 36, simplified.

Pick "At least one 6", then press Show the complement. Count the blue cells. Can you see why $1 - \frac{25}{36}$ is quicker than counting the orange ones? Then choose "Tap your own" and build the event "the red die is bigger than the blue die".

Example 2●●

Two fair dice are rolled. Find the probability that the total is 8.

$n(S) = 6 \times 6 = 36$

Each die has 6 faces and every pair is equally likely.

$(2,6), (3,5), (4,4), (5,3), (6,2)$, so $n(E) = 5$.

List the ordered pairs (red, blue) that add to 8. (2, 6) and (6, 2) are different cells.

$P(\text{total } 8) = \frac{5}{36}$

Divide favourable by total. 5 and 36 share no factor, so it can't simplify.

Complementary events

The complement of an event $A$ is everything in the sample space that is not in $A$. We write it as $\overline{A}$, $A'$ or "not $A$". An event and its complement can't both happen, and one of them must happen. Together they fill the whole sample space, so their probabilities add to 1.

Complementary events

$$P(\overline{A}) = 1 - P(A) \qquad\text{and}\qquad P(A) + P(\overline{A}) = 1$$

Example 3●●

Two dice are rolled. Find the probability of getting at least one 6.

$P(\text{at least one 6}) = 1 - P(\text{no 6})$

'At least one 6' means one 6 or two 6s, which is fiddly to count. Its complement is 'no 6 at all'.

$P(\text{no 6}) = \frac{5 \times 5}{36} = \frac{25}{36}$

With no 6, each die can show only 1 to 5, so there are 5 × 5 cells.

$P(\text{at least one 6}) = \frac{36}{36} - \frac{25}{36} = \frac{11}{36}$

Subtract from 1, writing 1 as 36/36.

Check it on the grid: the cells with a 6 are the bottom row and the right-hand column. That is $6 + 6 = 12$, minus 1 because (6, 6) was counted twice, giving 11.

Your turn

A jar of jelly beans holds 8 red, 5 green and 7 yellow. One is taken without looking. Find $P(\text{green})$ and $P(\text{not green})$.

Show the answer

There are $8 + 5 + 7 = 20$ jelly beans. $P(\text{green}) = \frac{5}{20} = \frac{1}{4}$. The complement is $P(\text{not green}) = 1 - \frac{1}{4} = \frac{3}{4}$. Check by counting: $8 + 7 = 15$ are not green, and $\frac{15}{20} = \frac{3}{4}$.

Flashcards

Probability with equally likely outcomes

$P(E) = \frac{n(E)}{n(S)}$: favourable outcomes over total outcomes.

Complement rule

$P(\text{not } A) = 1 - P(A)$

Signal to use the complement

"At least one": work out $1 - P(\text{none})$.

Practice

A class of 30 students has 9 who walk to school. One student is chosen at random. Find the probability, as a decimal, that the student walks.
Two fair dice are rolled. What is the probability of a double?
The probability that a particular Sydney Trains service is late is $0.15$. What is the probability, as a decimal, that it is not late?
Two dice are rolled. Which total is most likely?
Two dice are rolled. How many of the 36 outcomes give a total of 10 or more?
A barrel holds 50 numbered balls and one is drawn. $P(\text{the number is a multiple of 5}) = \frac{10}{50}$. What is $P(\text{not a multiple of 5})$?
Two dice are rolled. Find, as a decimal, the probability that the product of the two numbers is odd.
Which pair of events are complementary when one die is rolled?
Exam style: A family has two children. Assuming boys and girls are equally likely, find, as a decimal, the probability that they have at least one girl.
Exam style: Two dice are rolled. Mia says $P(\text{total of } 4) = \frac{1}{11}$ "because there are 11 possible totals from 2 to 12". What is the correct probability?