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

28.6 Conditional probability: the multiplication law 921
(a) Consider 1000 components. Then 650 are manufactured by machine M, 350 by
machineN.Ofthe650manufacturedbymachineM,95%willbeacceptable,that
is650× 95
100 = 617.5.Ofthe350manufacturedbymachineN,83%willbeacceptable,thatis350×
83
100 = 290.5.Soonaveragein1000components,617.5+
290.5 = 908 will be acceptable, thatisP(C) = 0.908.
(b) From part (a) we know that out of 1000 components 617.5 will be made by
machine M and be of an acceptable standard. Hence P(A ∩C) = 617.5
1000 =
0.6175.
(c) We requireP(C|A). The probability a component is acceptable given it is manufactured
by machine M is
P(C|A) = 0.95
(d) Machine M makes 65%of components, thatisP(A) = 0.65.
(e) We requireP(A|C). Consider again the 1000 components. On average 908 are
acceptable.Ofthese908acceptablecomponents,617.5aremanufacturedbymachine
M and 290.5 by machine N. We are told the component is acceptable and
sowemustrestrictattentiontothe908acceptablecomponents.So,outof908acceptablecomponents,617.5aremadebymachineM,thatisP(A|C)
= 617.5
908 =
0.68.
(f) We require P(B|C). Consider 1000 components. Machine M manufactures
650 components of which 617.5 are acceptable and hence 32.5 are unacceptable.
Machine N manufactures 350 components of which 290.5 are acceptable
and59.5areunacceptable.Thereare92unacceptablecomponentsofwhich59.5
weremadebymachineN.Theprobabilityofthecomponentbeingmadebymachine
Ngiven itisunacceptable is
P(B|C) = 59.5
92 = 0.647
that is, almost 65% of unacceptable components are manufactured by
machine N.
As an alternative method of solution, we can represent the information via a tree
diagram. This isshown inFigure 28.6.
617.5 (acceptable)
M
650
32.5 (unacceptable)
1000
N
350
290.5 (acceptable)
59.5 (unacceptable)
Figure28.6
Tree diagram forEngineeringapplication
28.12.
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