082-Engineering-Mathematics-Anthony-Croft-Robert-Davison-Martin-Hargreaves-James-Flint-Edisi-5-2017

28.6 Conditional probability: the multiplication law 919
the redundancy of a data stream to be reduced. The design of such codes is known as
coding theory. One complication is that most streams of data are not transmitted with
100%accuracyasaresultofthepresenceofnoisewithinthecommunicationchannel.It
isoftennecessarytobuildextraredundancyintoacodeinordertorecovertheseerrors.
EXERCISES28.5
1 Asourcegenerates sixcharacters, A, B,C,D, E, F,
with respective probabilities 0.05, 0.1, 0.25, 0.3, 0.15,
0.15.Calculate the average information percharacter
andthe redundancy.
2 Avisual displayunit has aresolutionof600 rowsby
800 columns.Ten different grey levelsare associated
with each pixel andtheir probabilitiesare 0.05,0.07,
0.09,0.10,0.11,0.13,0.12,0.12,0.11,0.10.Calculate
theaverageinformationcontentineachpictureframe.
3 Asourcegenerates five characters A, B,C,DandE
with respective probabilities of0.1, 0.15,0.2, 0.25
and0.3.
(a)Calculate the informationassociated with the
character B.
(b)Calculate the entropy.
(c)Calculate the redundancy.
4 Adata stream comprisesthe characters A, B,C,D
andEwithrespectiveprobabilitiesof0.23,0.16,0.11,
0.37 and 0.13.
(a) Calculate the informationassociated with the
character B.
(b) Calculate the informationassociated with the
character D.
(c) Calculate the entropy.
(d) Calculate the redundancy.
5 A data stream comprisesthe characters A, B,C,D
andEwithrespectiveprobabilitiesof0.12,0.21,0.07,
0.31 and 0.29.
(a) Which character carries the greatest information
content?
(b) Which character carries the leastinformation
content?
(c) Calculate the informationassociated with the
letter D.
(d) Calculate the entropy.
(e) Calculate the redundancy.
Solutions
1 2.3905,0.0752
2 3.2790
3 (a) 2.7370 (b) 2.2282 (c) 0.0404
4 (a) 2.6439 (b) 1.4344 (c) 2.1743
(d) 0.0636
5 (a) C (b) D (c) 1.6897
(d) 2.1501 (e) 0.0740
28.6 CONDITIONALPROBABILITY:THEMULTIPLICATIONLAW
Suppose two machines, M and N, both manufacture components. Of the components
made by machine M, 92% are of an acceptable standard and 8% are rejected. For machine
N, only 80% are of an acceptable standard and 20% are rejected. Consider now
the eventE:
E: a component isof an acceptable standard
If all the components are manufactured by machine M then P(E) = 0.92. However,
if all the components are manufactured by machine N then P(E) = 0.8. If half the
componentsaremanufacturedbymachineMandhalfbymachineNthenP(E) = 0.86.
Toseewhythisissoconsider1000components.OfthehalfmadebymachineM,92%×