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A survey asks two questions of 300 students: Do you live on campus? Are you satisfied with the recreational facilities in the student center?Answers to the latter question were categorized “reasonably satisfied,” “unsatisfied,” or “no definite opinion.” Let C be the event “on campus;” O be the event “off campus;” S be the event “reasonably satisfied;” U be the event ”unsatisfied;” and N be the event “no definite opinion.” Data are shown in the following table.
Survey Data
S | U | N | |
C | 127 | 31 | 42 |
O | 46 | 43 | 11 |
If an individual is selected on an equally likely basis from this group of 300, the probability of any of the events is taken to be the relative frequency of respondents ineach category corresponding to an event. There are 200 on campus members in the population, so $P\left(C\right)=200/300$ and $P\left(O\right)=100/300$ . The probability that a student selected is on campus and satisfied is taken to be $P\left(CS\right)=127/300$ . The probability a student is either on campus and satisfied or off campus and not satisfied is
If there is reason to believe that the population sampled is representative of the entire student body, then the same probabilities would be applied to any studentselected at random from the entire student body.
It is fortunate that we do not have to declare a single position to be the “correct” viewpoint and interpretation.The formal model is consistent with any of the views set forth. We are free in any situation to make the interpretation most meaningful and natural to the problem at hand . It is not necessary to fit all problems into one conceptual mold; nor is itnecessary to change mathematical model each time a different point of view seems appropriate.
Often we find it convenient to work with a ratio of probabilities. If A and B are events with positive probability the odds favoring A over B is the probability ratio $P\left(A\right)/P\left(B\right)$ . If not otherwise specified, B is taken to be A ^{c} and we speak of the odds favoring A
This expression may be solved algebraically to determine the probability from the odds
In particular, if $O\left(A\right)=a/b$ then $P\left(A\right)=\frac{a/b}{1+a/b}=\frac{a}{a+b}$ .
$O\left(A\right)=0.7/0.3=7/3$ . If the odds favoring A is 5/3, then $P\left(A\right)=5/(5+3)=5/8$ .
The countable additivity property (P3) places a premium on appropriate partitioning of events.
Definition. A partition is a mutually exclusive class
A partition of event A is a mutually exclusive class
Remarks .
We may begin with a sequence $\{{A}_{1}:1\le i\}$ and determine a mutually exclusive (disjoint) sequence $\{{B}_{1}:1\le i\}$ as follows:
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