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

928 Chapter 28 Probability
P(atleastone component isfaulty) =P(exactly one component isfaulty)
+P(both components arefaulty)
= 0.0768 +0.0016
= 0.0784
Asanalternativewecannotethatthecomplementof‘atleastoneofthecomponentsisfaulty’is‘noneofthecomponentsisfaulty’.Theprobabilitythatneither
of the components is faulty isgiven in(b). Hence
P(atleastone component isfaulty)
= 1 −P(none of the components isfaulty)
= 1 −0.9216
= 0.0784
(f) At least one of the components is not faulty means that one or both of the components
isnotfaulty. So
P(atleastone component isnotfaulty)
=P(exactly one component isnot faulty)
+P(both components are notfaulty)
= 0.0768 +0.9216
= 0.9984
Asanalternativewenotethatthecomplementof‘atleastoneofthecomponents
is not faulty’ is ‘none of the components are not faulty’. The last statement is
equivalent to‘both the components arefaulty’.Hence
P(atleastone component isnot faulty) = 1−P(both components arefaulty)
= 1 −0.0016
= 0.9984
Engineeringapplication28.16
Probabilityofacceptablecomponents
Machines 1, 2 and 3 manufacture resistors A, B and C, respectively. The probabilities
of their respective acceptabilities are 0.9, 0.93 and 0.81. One of each resistor is
selected atrandom.
(a) Find the probability thatthey are all acceptable.
(b) Find the probability thatatleastone resistor isacceptable.
Solution
Define eventsE 1
,E 2
andE 3
by
E 1
: resistor Ais acceptable P(E 1
) = 0.9
E 2
: resistor Bisacceptable P(E 2
) = 0.93
E 3
: resistor Cisacceptable P(E 3
) = 0.81