Elementary and Conditional Probability (Senior 5 Mathematics): free lessons | Ganzaa

Elementary and Conditional Probability

Lessons

Senior 5 · Mathematics

Elementary and Conditional Probability

4 lessons

Lesson 1 of 4

Experiments, sample spaces and the rules of probability

40 min

Thirty-six ways for two dice to land

In a board game you need a total of 7 on two dice to win. Is 7 more likely than 12? In S3 you found probabilities by counting regions of Venn diagrams. At A-level we make the ideas exact: list every possible result of an experiment, describe events as sets of results, and derive the rules that every probability obeys. Those rules then carry us to independence, conditional probability and Bayes' theorem, which doctors, engineers and insurers use every day.

Definition

A random experiment is an action whose result cannot be predicted with certainty, but whose possible results are known: tossing a coin, rolling two dice, choosing a pupil at random.
Each possible result is an outcome. The set of all outcomes is the sample space, written $\Omega$ (or $S$ ); $n(\Omega)$ is the number of outcomes.
An event is a subset of $\Omega$ . It occurs when the outcome belongs to it. A single outcome $\{\omega\}$ is an elementary event; $\Omega$ is the sure event and $\varnothing$ the impossible event.
The complement $A^{'}$ (or $\bar{A}$ ) is " $A$ does not occur"; $A \cup B$ is " $A$ or $B$ (or both)"; $A \cap B$ is " $A$ and $B$ ".

Example: tossing three coins, $\Omega = \{HHH, HHT, HTH, THH, HTT, THT, TTH, TTT\}$ , $n(\Omega) = 8$ , and "exactly two heads" is the event $\{HHT, HTH, THH\}$ .

Important

Axioms of probability. A probability on a finite sample space $\Omega$ gives each event $A$ a number $P (A)$ such that

$P(A) \ge 0$ for every event $A$ ;
$P(\Omega) = 1$ ;
if $A \cap B = \varnothing$ , then $P(A \cup B) = P(A) + P(B)$ .

Consequences. $P (A^{'}) = 1 - P (A)$ ; $P(\varnothing) = 0$ ; if $A \subseteq B$ then $P(A) \le P(B)$ ; and $0 \le P(A) \le 1$ .

Reason for the complement rule. $A$ and $A^{'}$ do not overlap and together make $\Omega$ , so by axioms 3 and 2, $P(A) + P(A') = P(\Omega) = 1$ . Taking $A = \Omega$ gives $P(\varnothing) = 0$ .

Equally likely outcomes. If all $n(\Omega)$ outcomes are equally likely (a fair die, a random choice), then

P(A) = \frac{n(A)}{n(\Omega)} = \frac{\text{number of outcomes in } A}{\text{number of possible outcomes}} .

The sample space of two dice

Colour the dice red and blue so that "red 2, blue 5" and "red 5, blue 2" are different outcomes. Then all 36 outcomes are equally likely.

Worked example 1: two dice

Two fair dice are rolled. Find the probability that (a) the total is 7; (b) the total is at least 10; (c) at least one die shows a 6.

Solution. $n(\Omega) = 36$ . (a) The orange diagonal has 6 cells: $P = \frac{6}{36} = \frac{1}{6}$ . (b) Totals 10, 11, 12 occur in $3 + 2 + 1 = 6$ cells: $P = \frac{1}{6}$ . So a total of 7 is exactly as likely as "10 or more", and six times as likely as 12 ( $\frac{1}{36}$ ).

(c) The complement is "no 6 on either die", which has $5 \times 5 = 25$ outcomes. So $P = 1 - \frac{25}{36} = \frac{11}{36}$ .

Definition

Counting choices. The number of ways to choose $r$ objects from $n$ different objects, when order does not matter, is the number of combinations

\binom{n}{r} = {}^{n}C_{r} = \frac{n!}{r!\,(n - r)!}, \qquad n! = n(n - 1)\cdots 2 \cdot 1,\ 0! = 1 .

Example: $\binom{5}{2} = \frac{5 \times 4}{2 \times 1} = 10$ ways to choose 2 pupils from 5. When order matters, there are $\frac{n!}{(n - r)!}$ arrangements.

Worked example 2: choosing a team at random

Three pupils are chosen at random from 6 girls and 4 boys to represent the class at a district science fair. Find the probability that (a) all three are girls; (b) at least one is a boy; (c) exactly two are girls.

Solution. $n(\Omega) = \binom{10}{3} = 120$ , all equally likely. (a) $\binom{6}{3} = 20$ , so $P = \frac{20}{120} = \frac{1}{6}$ . (b) By the complement rule, $P = 1 - \frac{1}{6} = \frac{5}{6}$ . (c) Choose 2 of the 6 girls and 1 of the 4 boys: $\binom{6}{2}\binom{4}{1} = 15 \times 4 = 60$ , so $P = \frac{60}{120} = \frac{1}{2}$ .

Worked example 3: quality control

A box of 12 bulbs contains 3 defective ones. Two bulbs are taken at random. Find the probability that both work, and that exactly one is defective.

Solution. $n(\Omega) = \binom{12}{2} = 66$ . Both work: $\binom{9}{2} = 36$ , so $P = \frac{36}{66} = \frac{6}{11}$ . Exactly one defective: $\binom{9}{1}\binom{3}{1} = 27$ , so $P = \frac{27}{66} = \frac{9}{22}$ . (Both defective: $\frac{3}{66} = \frac{1}{22}$ , and $\frac{12}{22} + \frac{9}{22} + \frac{1}{22} = 1$ . ✓)

Warning

Check that outcomes really are equally likely. The totals 2 to 12 of two dice are 11 outcomes, but not equally likely, so $P(\text{total } 7) \ne \frac{1}{11}$ .
"At least one" is usually easiest through the complement "none".
A probability is never negative and never more than 1. An answer outside $[0, 1]$ is an error.

Tip

Make the dice different colours in your mind, even if they are not. It turns 21 unequal "unordered" results into 36 equally likely ordered ones.

FuseSchool: Sample Spaces (about 3 min). A short summary of listing every outcome, including the two-dice grid used in this lesson.

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Practice

Questions 1 to 3 are routine; 4 to 6 need more care; 7 asks for a reason.

Write the sample space for tossing a coin and rolling a die. How many outcomes are there?
Two fair dice are rolled. Find $P(\text{total} = 5)$ and $P(\text{a double})$ .
$P (A) = 0.35$ . Find $P (A^{'})$ .
A committee of 4 is chosen at random from 5 men and 7 women. Find the probability that it has exactly 2 women.
Find the probability that a 4-digit code, each digit chosen at random from 0 to 9, has no repeated digit.
Three fair coins are tossed. Find the probability of at least one head, and of more heads than tails.
Explain why $P(A) \le P(B)$ whenever $A \subseteq B$ . (Hint: write $B$ as the union of $A$ and $B \cap A'$ .)

Show answersHide answers

$\{H1, H2, \dots, H6, T1, \dots, T6\}$ : 12 outcomes.
Totals of 5: $(1, 4), (2, 3), (3, 2), (4, 1)$ , so $\frac{4}{36} = \frac{1}{9}$ . Doubles: $\frac{6}{36} = \frac{1}{6}$ .
$0.65$ .
$\frac{\binom{7}{2}\binom{5}{2}}{\binom{12}{4}} = \frac{21 \times 10}{495} = \frac{210}{495} = \frac{14}{33}$ .
$\frac{10 \times 9 \times 8 \times 7}{10^4} = \frac{5040}{10000} = 0.504$ .
$1 - \frac{1}{8} = \frac{7}{8}$ ; two or three heads: $\frac{3 + 1}{8} = \frac{1}{2}$ .
$A$ and $B \cap A'$ do not overlap and their union is $B$ , so $P(B) = P(A) + P(B \cap A') \ge P(A)$ , since $P(B \cap A') \ge 0$ .

Sources and credits

Written by Ganzaa: the explanations, the sample-space grid, the worked examples, the practice items and the quiz questions. Every probability was checked by counting with a computer.
Video: Sample Spaces | Statistics & Probability | Maths, FuseSchool – Global Education, embedded from YouTube.

1Thirty-six ways for two dice to land

2The sample space of two dice

Thirty-six ways for two dice to land

The sample space of two dice