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] $$\,P(A\cup B)=P(A)+P(B)\,$$
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] $$\,P(A\cup B)=P(A)+P(B)\,$$
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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 $\,2\cdot 3 = 6\,$. |
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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?
$\,2\,$ 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.
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
by practicing the exercise at the bottom of this page.
When you're done practicing, move on to:
Probability Tree Diagrams