MORE PROBABILITY CONCEPTS

LESSON READ-THROUGH
by Dr. Carol JVF Burns (website creator)
Follow along with the highlighted text while you listen!

• PRACTICE (online exercises and printable worksheets)
• Need an introduction to probability first?
  Basic Probability Concepts

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 $\,2\cdot 3 = 6\,$.

On this exercise, you will not key in your answer. However, you can check to see if your answer is correct.

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