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Empirical probability based on survey data

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

Survey data
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 ( C ) = 200 / 300 and P ( O ) = 100 / 300 . The probability that a student selected is on campus and satisfied is taken to be P ( C S ) = 127 / 300 . The probability a student is either on campus and satisfied or off campus and not satisfied is

P ( C S O U ) = P ( C S ) + P ( O U ) = 127 / 300 + 43 / 300 = 170 / 300

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.

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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.

Probability and odds

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 ( A ) / P ( B ) . If not otherwise specified, B is taken to be A c and we speak of the odds favoring A

O ( A ) = P ( A ) P ( A c ) = P ( A ) 1 - P ( A )

This expression may be solved algebraically to determine the probability from the odds

P ( A ) = O ( A ) 1 + O ( A )

In particular, if O ( A ) = a / b then P ( A ) = a / b 1 + a / b = a a + b .

O ( A ) = 0 . 7 / 0 . 3 = 7 / 3 . If the odds favoring A is 5/3, then P ( A ) = 5 / ( 5 + 3 ) = 5 / 8 .

Partitions and boolean combinations of events

The countable additivity property (P3) places a premium on appropriate partitioning of events.

Definition. A partition is a mutually exclusive class

{ A i : i J } such that Ω = i J A i

A partition of event A is a mutually exclusive class

{ A i : i J } such that A = i J A i

Remarks .

  • A partition is a mutually exclusive class of events such that one (and only one) must occur on each trial.
  • A partition of event A is a mutually exclusive class of events such that A occurs iff one (and only one) of the A i occurs.
  • A partition (no qualifier) is taken to be a partition of the sure event Ω .
  • If class { B i : ı J } is mutually exclusive and A B = i J B i , then the class { A B i : ı J } is a partition of A and A = i J A B i .

We may begin with a sequence { A 1 : 1 i } and determine a mutually exclusive (disjoint) sequence { B 1 : 1 i } as follows:

Questions & Answers

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The nanotechnology is as new science, to scale nanometric
nanotechnology is the study, desing, synthesis, manipulation and application of materials and functional systems through control of matter at nanoscale
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A fair die is tossed 180 times. Find the probability P that the face 6 will appear between 29 and 32 times inclusive
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Source:  OpenStax, Applied probability. OpenStax CNX. Aug 31, 2009 Download for free at http://cnx.org/content/col10708/1.6
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