The outcomes of a chance event such as tossing a coin, rolling a die, or drawing a card from a deck can be describe by two related methodologies,

- This concerns simultaneously tossing a fair coin while rolling a die with six
numbered faces. The faces are numbered one through six and the die is
also fair.
- Draw the sample space.
- How big is the sample space in terms of points corresponding to die-roll/coin-toss outcomes?
- What is the probability of a three(die)-head(coin) outcome?

- Four births occur. Of the following two events which is most probable?
- Half male, half female.
- Exactly three of one sex.

The contrast with analysis is provided by

If two random things take place that have nothing to do with one another there are three separate probabilities to consider: that of the first alone, of just the second, and that from both taking place. When they truly have no relationship, the last probability is the product of the first two, and those are called

Dependence of two random events called A and B is when the product of the individual probabilities P(A) times P(B) does not create P(A, B), the probability of both occurring simultaneously. When we know that one of the events has occurred, say B, we can find the way that impacts the probability of A taking place. We write this as P(A|B) (read "probability of A given that B occurred" or "the conditional probability of A given B"). The basis of a decision-making is the following definition of

Multiplication by the denominator portion P(B) and But (1) holds whatever the letter labels are used. So by interchange of A and B, followed by clearing away division by P(A), in the similar expression, leads to the following:

Suppose a ten sound string is associated with a word like

In many situations there is neither independence nor complete dependence. In analytical terms, this middle ground means P(A, B) differs from P(A)P(B) but the parts of A differ in their influence on how it is related to B.

Independence means that (1) becomes P(A|B) = P(A, B) = P(A)P(B)/P(B) = P(A) and likewise:

Can (2) deal with the tenth-sound-in-a-word decision? Formally,

If the current sound value depends only on the immediately prior one, this becomes:

If we know: 1) for every possible s

A sometimes reasonable assumption is that

4/22/02 Version |
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©2002 Allen Klinger |