Events and the rules of probability

The sample space is the set of all the possible outcomes of an experiment. It is written with the letter $\displaystyle U$. An event is a part of the sample space – a set of outcomes.

A die is thrown once:

• The sample space: $\displaystyle {U}={\{1,2,3,4,5,6\}}$.
• The event $\displaystyle A$ – “an even number comes up”: $\displaystyle {A}={\{2,4,6\}}$.
• The event $\displaystyle B$ – “a number greater than 3 comes up”: $\displaystyle {B}={\{4,5,6\}}$.

Two special events:

• The certain event – the whole sample space, $\displaystyle U$: “a number from 1 to 6 comes up” always happens.
• The impossible event – the empty set, $\displaystyle \emptyset$: “a 7 comes up” never happens.

From two events we build new ones:

• The union $\displaystyle A\cup B$ – “$\displaystyle A$ or $\displaystyle B$”: the outcomes in at least one of them, both included. Here $\displaystyle {A\cup B}={\{2,4,5,6\}}$.
• The intersection $\displaystyle A\cap B$ – “$\displaystyle A$ and $\displaystyle B$”: the outcomes in both. Here $\displaystyle {A\cap B}={\{4,6\}}$.

In a Venn diagram the rectangle is the sample space, and each circle is an event. The part the two circles share is the intersection, and the two circles together are the union.

The outcomes 1 and 3 are in the rectangle and outside the circles: they are neither in $\displaystyle A$ nor in $\displaystyle B$.

Two events are disjoint when they have no outcome in common: $\displaystyle {A\cap B}={\emptyset}$. They cannot happen together. In a Venn diagram their circles do not touch.

• When a die is thrown, “a 1 comes up” and “an even number comes up” are disjoint.
• “An even number comes up” and “a number greater than 3 comes up” are not disjoint: 4 and 6 are in both.

The complement of $\displaystyle A$, written $\displaystyle \overline{A}$, is all the outcomes not in $\displaystyle A$: “$\displaystyle A$ does not happen”.

• $\displaystyle {A}={\{2,4,6\}}$, so $\displaystyle {\overline{A}}={\{1,3,5\}}$.
• An event and its complement are disjoint, and their union is the whole sample space: $\displaystyle {A\cup\overline{A}}={U}$.

The probability of an event is a number from 0 to 1 that says how likely the event is to happen. It is worked out in one of three ways, by what is known about the experiment:

• Equally likely outcomes – the number of outcomes in the event divided by the number of outcomes in the sample space. For the die: $\displaystyle {P(A)}={\frac{3}{6}}={\frac{1}{2}}$.
• Relative frequency – when the outcomes are not equally likely, the experiment is repeated many times. A drawing pin thrown 500 times landed on its head 320 times: $\displaystyle {P\approx\frac{320}{500}}={0.64}$.
• A ratio of areas – when a point is chosen at random inside a shape, the probability of hitting a part of it is the part’s area divided by the whole shape’s area.

A dart hits a square board of side 40 cm at random.
In the middle of the board is a red square of side 20 cm.
$\displaystyle R$ – the event “the dart hits the red square”.

Some rules follow from the definition of probability, for every event $\displaystyle A$ and every event $\displaystyle B$:

• $\displaystyle {P(U)}={1}$: the certain event; $\displaystyle {P(\emptyset)}={0}$: the impossible event.
• If all the outcomes of $\displaystyle A$ are also in $\displaystyle B$ (written $\displaystyle A\subseteq B$), then $\displaystyle P(A)\le P(B)$. “A 6 comes up” lies inside “an even number comes up”, so $\displaystyle \frac{1}{6}\le\frac{1}{2}$.

The complement: $\displaystyle A$ and $\displaystyle \overline{A}$ together are the whole sample space, so their probabilities add up to 1.

A union of disjoint events – the probabilities add up, since no outcome is counted twice:

So too for several events, disjoint from one another: $\displaystyle {P(A_1\cup A_2\cup\dots\cup A_n)}={P(A_1)+P(A_2)+\dots+P(A_n)}$.

When the events are not disjoint, $\displaystyle P(A)+P(B)$ counts the intersection twice – once in $\displaystyle A$ and once in $\displaystyle B$. So it is taken away once:

For the die: $\displaystyle {P(A)}={\frac{3}{6}}$, $\displaystyle {P(B)}={\frac{3}{6}}$, $\displaystyle {P(A\cap B)}={\frac{2}{6}}$.

This agrees with counting: $\displaystyle {A\cup B}={\{2,4,5,6\}}$ – four outcomes out of six.

A die is thrown once.
$\displaystyle C$ – “a number less than 3 comes up”.
$\displaystyle D$ – “an odd number comes up”.

Write $\displaystyle C$, $\displaystyle D$, $\displaystyle C\cup D$ and $\displaystyle C\cap D$ as sets of outcomes.

Work out $\displaystyle P(C\cup D)$.

Are $\displaystyle C$ and $\displaystyle D$ disjoint events? Explain.

Two dice are thrown.
$\displaystyle A$ – “the sum of the numbers is 7”.
$\displaystyle B$ – “a 6 comes up on at least one of the dice”.

Mark in the table the outcomes of the event $\displaystyle A\cup B$.

Work out $\displaystyle P(A\cup B)$ from the table.

$\displaystyle A$ has 6 outcomes, $\displaystyle B$ has 11, and $\displaystyle A\cap B$ has 2. Work out $\displaystyle P(A\cup B)$ by the union formula.

A student at a school is chosen at random.
The probability that he plays an instrument is $\displaystyle 0.5$.
The probability that he sings in the choir is $\displaystyle 0.4$.
The probability that he both plays and sings is $\displaystyle 0.15$.

What is the probability that the student plays an instrument or sings in the choir?

What is the probability that the student neither plays an instrument nor sings in the choir?

$\displaystyle A$ and $\displaystyle B$ are disjoint events.
$\displaystyle {P(A)}={0.3}$, $\displaystyle {P(B)}={0.45}$.

Work out $\displaystyle P(A\cup B)$.

Ron says: “There are two disjoint events whose probabilities are $\displaystyle 0.7$ and $\displaystyle 0.6$.” Explain why this is impossible.

A dart hits a square board of side 50 cm at random.
In the middle of the board is a circle of radius 10 cm.

What is the probability that the dart hits the circle? Round to two decimal places.