The sample space and events

Probability, grade 11 (5 units) Lesson 1 of 16Learning

The sample space and events

The sample space

An experiment in probability is an action whose result is not known in advance: throwing a die, choosing a student at random, a football match. The sample space is the set of all the possible outcomes of the experiment. It is written $\displaystyle U$ .

A die is thrown: $\displaystyle {U}={\{1,2,3,4,5,6\}}$ .
A coin is tossed twice: four outcomes – heads-heads, heads-tails, tails-heads, tails-tails. The outcome heads-tails – heads on the first toss and tails on the second – is not the outcome tails-heads.
A football team plays a match: three outcomes – a win, a draw, a loss.

In this chapter the sample space is always finite: it has a certain number of outcomes.

Simple and compound events

An event is a set of outcomes taken from the sample space. The event happened if the experiment’s outcome is one of its outcomes.

A simple event – one outcome: “a 5 comes up”, $\displaystyle \{5\}$ .
A compound event – several outcomes. It is described with words like “or”, “greater than”, “at most”, “at least”:
- “an even number greater than 3 comes up”: $\displaystyle \{4,6\}$ ;
- “at most 2 comes up” – 2 or less: $\displaystyle \{1,2\}$ ;
- “at least 5 comes up” – 5 or more: $\displaystyle \{5,6\}$ .
The certain event – the whole sample space, $\displaystyle U$ : “a number less than 10 comes up” always happens.
The impossible event – the empty set, $\displaystyle \emptyset$ : “a 7 comes up” never happens.

Equally likely outcomes

When all the outcomes in the sample space have the same probability – a fair die, a fair coin, a choice at random – the probability of an event is the share of its outcomes among all the outcomes:

\displaystyle P(A)=\frac{k}{n}

$\displaystyle k$ – the number of outcomes in the event $\displaystyle A$ ; $\displaystyle n$ – the number of outcomes in the sample space.

A fair die is thrown. $\displaystyle A$ – “an even number greater than 3 comes up”.

P(A)=\frac{2}{6}=\frac{1}{3}

It follows that $\displaystyle {P(U)}={\frac{n}{n}}={1}$ , that $\displaystyle {P(\emptyset)}={\frac{0}{n}}={0}$ , and that the probability of every event is between 0 and 1.

The sample space of two dice

In an experiment of two stages each outcome is a pair. A blue die and a red die are thrown, so the sample space has $\displaystyle {6\cdot6}={36}$ equally likely pairs – a square for each pair in the outcome table. Each square holds the sum of the two dice.

$\displaystyle B$ – “the sum of the two dice is at most 4”. The squares of $\displaystyle B$ are marked:

$\displaystyle B$ holds 6 pairs out of 36:

P(B)=\frac{6}{36}=\frac{1}{6}

A sample space whose outcomes are not equally likely

The sample space of an experiment can be written in several ways, but the formula $\displaystyle {P(A)}={\frac{k}{n}}$ holds only when the outcomes are equally likely.

If only the sum of the two dice is written down, the sample space is $\displaystyle \{2,3,4,\dots,12\}$ – 11 outcomes. But:

the sum 2 has one pair, (1,1): $\displaystyle {P(2)}={\frac{1}{36}}$ ;
the sum 7 has six pairs, from (1,6) to (6,1): $\displaystyle {P(7)}={\frac{6}{36}}$ .

The probability that the sum is 7 is not $\displaystyle \frac{1}{11}$ . To work it out, go back to the sample space of the 36 pairs, where all the outcomes are equally likely.

Exercise 1

A jar holds 10 slips of paper, numbered from 1 to 10.
One slip is taken out at random.
$\displaystyle A$ – “the number on the slip is even or greater than 7”.

Write the sample space and the event $\displaystyle A$ as sets of outcomes.

Work out $\displaystyle P(A)$ .

/ – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – check and a new line · * – times · ^ – power, ↓ – end of the power

/ – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – a new line · * – times · ^ – power, ↓ – end of the power

Exercise 2

A blue die and a red die are thrown.
Mark in the table the outcomes of the event “the sum of the two dice is at least 9”, and work out its probability.

/ – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – a new line · * – times · ^ – power, ↓ – end of the power

Exercise 3

A blue die and a red die are thrown.
Each square holds the difference between the larger number and the smaller one.
Mark the outcomes of the event “the difference between the two dice is at most 1”, and work out its probability.

/ – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – a new line · * – times · ^ – power, ↓ – end of the power

Exercise 4

A family has three children.
For each child, being a boy and being a girl are equally likely.

Write the sample space. Write each outcome in the order of the children from the eldest, for example BGB – a boy, then a girl, then a boy.

What is the probability that the family has exactly two girls?

/ – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – a new line · * – times · ^ – power, ↓ – end of the power

Dana says: “The number of girls in the family can be 0, 1, 2 or 3. These are four outcomes, so the probability of exactly two girls is $\displaystyle \frac{1}{4}$ .”
Explain Dana’s mistake.

Formulas: probability, grade 11 (5 units)

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The rules of probability

Equally likely outcomes – $\displaystyle k$ the outcomes in the event $\displaystyle A$ , $\displaystyle n$ the outcomes in the sample space
$\displaystyle P(A)=\frac{k}{n}$
Every event; $\displaystyle {P(\emptyset)}={0}$ , and $\displaystyle {P(U)}={1}$ for the sample space $\displaystyle U$
$\displaystyle 0\le P(A)\le1$
An event contained in another event
$\displaystyle P(A)\le P(B)$ $\displaystyle A\subseteq B$
The complement $\displaystyle \overline{A}$ – $\displaystyle A$ does not happen
$\displaystyle P(\overline{A})$ $\displaystyle =$ $\displaystyle 1-P(A)$
The union – $\displaystyle A$ or $\displaystyle B$ (or both); $\displaystyle A\cap B$ – the intersection, both $\displaystyle A$ and $\displaystyle B$
$\displaystyle P(A\cup B)$ $\displaystyle =$ $\displaystyle P(A)+P(B)-P(A\cap B)$
Disjoint events – no two of them have an outcome in common
$\displaystyle P(A_1\cup A_2\cup\dots\cup A_n)=P(A_1)+P(A_2)+\dots+P(A_n)$

Conditional probability and independence

The probability of $\displaystyle A$ given $\displaystyle B$
$\displaystyle P(A/B)=\frac{P(A\cap B)}{P(B)}$
A path in a tree – the product of the probabilities on its branches
$\displaystyle P(A\cap B)$ $\displaystyle =$ $\displaystyle P(B)\cdot P(A/B)$
Independent events
$\displaystyle P(A\cap B)$ $\displaystyle =$ $\displaystyle P(A)\cdot P(B)$
Independent events – knowing that $\displaystyle B$ happened does not change the probability of $\displaystyle A$
$\displaystyle P(A/B)$ $\displaystyle =$ $\displaystyle P(A)$
The law of total probability – the paths that end in $\displaystyle A$ , added up
$\displaystyle P(A)=P(B)\cdot P(A/B)+P(\overline{B})\cdot P(A/\overline{B})$

Counting and Bernoulli's formula

$\displaystyle n$ factorial; $\displaystyle {0!}={1}$
$\displaystyle n!=1\cdot2\cdot3\cdot\dots\cdot n$
Ordered selections – $\displaystyle k$ different objects out of $\displaystyle n$ , one after another
$\displaystyle n\cdot(n-1)\cdot\dots\cdot(n-k+1)=\frac{n!}{(n-k)!}$
The binomial coefficient – the ways to choose $\displaystyle k$ objects out of $\displaystyle n$ , order not counted
$\displaystyle \binom{n}{k}=\frac{n!}{k!(n-k)!}$
Bernoulli’s formula – the probability of $\displaystyle k$ successes in $\displaystyle n$ independent trials, with the probability of success $\displaystyle p$ in each
$\displaystyle P_n(k)=\binom{n}{k}p^k(1-p)^{n-k}$
At least one success in $\displaystyle n$ trials – by the complement
$\displaystyle 1-P_n(0)=1-(1-p)^n$