Properties of Probabilities
Recall that the probability of an event $\,E\,$
is denoted by
$\,P(E)\,$.

For any event
$\,E\,$, $\,0\le P(E)\le 1\,$.
That is, a probability is always a number between $\,0\,$ and $\,1\,$.

If $\,S\,$ is the sample space,
then $\,P(S)=1\,$.
That is, there's a $\,100\%\,$ chance that something will happen.

For a finite sample space (i.e., an experiment with a finite number of outcomes):
a probability of $\,1\,$ means the event will definitely occur;
a probability of $\,0\,$ means the event will definitely not occur.
Notice that if the experiment is to randomly choose a real number from the interval $\,[0,1]\,$,
then the probability of choosing (say) the number $\,0.5\,$ is zero,
and yet this outcome could occur.
This is why a finite sample space is needed in the above statement.

For any events $\,A\,$ and $\,B\,$:
$$\cssId{s17}{P(A\cup B)}
\cssId{s18}{=P(A)+P(B)P(A\cap B)}$$
If you were just to add the two probabilities,
then the intersection gets added twice;
this extra probability must be subtracted.

Recall that
$\,\overline{E}\,$ denotes the complement of $\,E\,$.
For any event, either it happens, or it doesn't!
Therefore, $\,P(E)+P(\overline{E})=1\,$.
Equivalently, $\,P(\overline{E})=1 P(E)\,$.

Recall that
$\,\emptyset \,$
denotes the empty set.
Events $\,A\,$ and $\,B\,$ are said to be
mutually exclusive if
$\,A\cap B = \emptyset\,$.
That is, two events are mutually exclusive when they have nothing in common
(their intersection is empty).
Since $\,P(\emptyset) = 0\,$, for mutually exclusive events, the above formula is simpler:
$$\cssId{s33}{\,P(A\cup B)=P(A)+P(B)}\,$$
MULTIPLICATION COUNTING PRINCIPLE
If there are $\,F\,$ choices for how to perform a first act,
and for each of these $\,F\,$ ways,
there are $\,S\,$ choices for how to perform a second act,
then there are $\,F\cdot S\,$ ways to perform the acts in succession.
(The idea extends to more than $\,2\,$ acts.)
The idea is illustrated by the diagram at right.
If there are $\,2\,$ piles, with $\,3\,$ in each pile,
then the total is $\,2\cdot 3 = 6\,$.


EXAMPLE (pizza choices)
Suppose that a pizza shop offers $\,3\,$ types of crust,
$\,2\,$ different types of cheese,
and $\,7\,$ different toppings.
A singletopping pizza consists of a choice of crust,
a choice of cheese (if desired),
and a choice of topping.
How many different singletopping pizzas are there?
Solution:
There are $\,3\,$ choices for the crust,
$\,3\,$ choices for cheese (none, first type, second type),
and $\,7\,$ choices for the single topping.
By the Multiplication Counting Principle,
there are $\,3\cdot 3\cdot 7 = 63\,$ possible singletopping pizzas.
Master the ideas from this section
by practicing the exercise at the bottom of this page.
When you're done practicing, move on to:
Probability Tree Diagrams