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

922 Chapter 28 Probability
(a) There are617.5 +290.5 = 908 acceptable components, so
P(component isacceptable) = 0.908
(b) There are617.5 components which are acceptable and made by machine M. So
P(component isacceptable and made by machine M)
=P(C ∩A) = 0.6175
(c) Of the 650 components made by machine M, 617.5 areacceptable, and so
P(component isacceptable given itismade by machine M) =P(C|A)
= 617.5
650 = 0.95
(d) Of the 1000 components, 650 aremade by machine M, so
P(component ismade by machine M) =P(A) = 0.65
(e) Thereare908acceptablecomponents.Ofthese,617.5aremadebymachineM.
Hence
P(component ismade by machine M given itisacceptable) =P(A|C)
= 617.5
908 = 0.68
(f) There are 32.5 + 59.5 = 92 unacceptable components. Of these 92, 59.5 are
made by machine N.Thus
P(component ismade by machine Ngiven itisnot acceptable) =P(B|C)
= 59.5
92 = 0.647
28.6.1 Themultiplicationlaw
Consider events A and B for which A ∩B ≠ ◦/ as shown in Figure 28.7. Suppose we
knowthateventAhasoccurredandweseektheprobabilitythatBoccurs,thatisP(B|A).
Knowing event A has occurred we can restrict our attention to the set A. Event B will
occur ifany outcome is inA ∩B. Hence,
E
P(B|A) =
P(A ∩B)
P(A)
A
B
Themultiplication lawofprobabilitystates:
P(A ∩B) =P(A)P(B|A)
A > B
Figure28.7
AandBare not
mutually exclusive.
Since the compound event ‘Aand B’ isidentical to‘B and A’ wemay alsosay
P(A ∩B) =P(B ∩A) =P(B)P(A|B)
Consider Engineering application 28.12(e). We require the probability that the component
was manufactured by machine M, given itisacceptable. This isP(A|C). Now