MORE PROBABILITY CONCEPTS
PROPERTIES OF PROBABILITIES
Recall that the probability of an event $\,E\,$ is denoted by [beautiful math coming... please be patient] $\,P(E)\,$.

For any event [beautiful math coming... please be patient] $\,E\,$, $\,0\le P(E)\le 1\,$.
That is, a probability is always a number between $\,0\,$ and $\,1\,$.

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 [beautiful math coming... please be patient] $\,[0,1]\,$,
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.

If [beautiful math coming... please be patient] $\,S\,$ is the sample space, then $\,P(S)=1\,$.
That is, there's a $\,100\%\,$ chance that something will happen.

For any events $\,A\,$ and $\,B\,$: [beautiful math coming... please be patient] $$P(A\cup B)=P(A)+P(B)-P(A\cap B)$$ If you were just to add the two probabilities, then the intersection gets added twice;
this extra probability must be subtracted.

Recall that [beautiful math coming... please be patient] $\,\overline{E}\,$ denotes the complement of $\,E\,$.
For any event, either it happens, or it doesn't!
Therefore, [beautiful math coming... please be patient] $\,P(E)+P(\overline{E})=1\,$.
Equivalently, [beautiful math coming... please be patient] $\,P(\overline{E})=1 -P(E)\,$.

Recall that $\,\emptyset \,$ denotes the empty set.
Events $\,A\,$ and $\,B\,$ are said to be mutually exclusive if [beautiful math coming... please be patient] $\,A\cap B = \emptyset\,$.
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: [beautiful math coming... please be patient] $$\,P(A\cup B)=P(A)+P(B)\,$$
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 single-topping pizza consists of a choice of crust,
a choice of cheese (if desired), and a choice of topping.

How many different single-topping 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 single-topping 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


On this exercise, you will not key in your answer.
However, you can check to see if your answer is correct.
PROBLEM TYPES:
1 2 3 4 5 6 7 8 9 10 11 12 13 14
AVAILABLE MASTERED IN PROGRESS

(MAX is 14; there are 14 different problem types.)