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Probabilities

We want the probability of event. Probabilities have to satisfy only the following three requirements:

A probability  is an assignment of numbers to events so that

1.
The probability of an event A is a number between 0 and 1.
2.
The probability of the sample space is 1.
3.
If two events cannot occur at the same time the probability that one or the other occurs is the sum of the probabilities of the individual events.

We will denote the probability of A by P(A). Notice that the first two requirements are not special. They simply state that probabilities are numbers between 0 and 1 and if the experiment is performed the sample space occurs with probability 1. The third requirement makes sense intuitively. For example, in the game of craps, the events A = sum of upfaces is 7 and B = sum of upfaces is 11 cannot occur at the same time, so the probability of a 7 or 11 is P(A) + P(B).

For a discrete (finite) sample space there is an easy way to obtain many examples of probabilities. Consider a subspace with m elements, say, $S = \{x_1, ..., x_m\}$. Let p1, ..., pm be m fractions between 0 and 1 which sum to 1. For an event A, consider the assignment

P(A) = sum of all pi's for which the element xi is in A.

Then the assignment P is a probability.


Example:  Suppose a spinner with the numbers 1 through 6 on it is spun and the number spun is observed. The sample space is S = {1, 2,3,4,5,6}. Let A be the event A = {1,2}. The following are four different probabilities on S and the resulting probability of A = {1,2} .

1.
p1 = p2 = p3 = p4 = p5 = p6 = 1/6
P(A)= 2/6 = 1/3.
2.
$p_1 = p_2 = .25, \, p_3 = p_4 = .15, \, p_5 = p_6 = .1$.
P(A)= .50.
3.
$p_1 = .10, \, p_2 = .25, \, p_3 = .1, \, p_4 = .15, \, p_5 = .30, \, p_6 = .1$.
P(A)= .35.
4.
$p_1 = 1/21, \, p_2 = 2/21, \, p_3 = 3/21,\, p_4 = 4/21, \, p_5 = 5/21, \, p_6 = 6/21$.
P(A)= 3/21= 1/7.
5.
$p_1 = .80, \, p_2 = .04, \, p_3 = .10, \, p_4 = .02, \, p_5 = .03, \, p_6 = .01$
P(A) = .84.
6.
$p_1 = 3/12, \, p_2 = 2/12, \, p_3 = 2/12, \, p_4 = 1/12, \, p_5 = 3/12, \, p_6 = 1/12$
P(A) = 5/12.


Exercise 3.2.1  
1.
In the last example, obtain 5 more different probabilities on S.
2.
In the last example, how many probabilities are there on S?
3.
In Exercise 2.1, #1, assume the spinner is fair. What is the assignment of probabilities? What is the probability of A, the event of interest? What is the probability of Ac, the event of interest?
4.
In Exercise 2.1, #3, assume that the probability of any topping on a pizza is 1/2. What is the assignment of probabilities? What is the probability of A, the event of interest? What is the probability of Ac, the event of interest?


next up previous contents index
Next: More on Probability Up: Probability Previous: Introduction

2001-01-01