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

28.7 Independent events 927
Solution
We define eventsF 1
andF 2
tobe
F 1
: the first component isfaulty
F 2
: the second component isfaulty
Then eventsF 1
andF 2
are
Then
F 1
: the first component isnotfaulty
F 2
: the second component isnotfaulty
P(F 1
) =P(F 2
) = 0.04
P(F 1
)=P(F 2
)=1−0.04=0.96
(a) We requireboth components tobefaulty:
P(F 1
∩F 2
) =P(F 1
)P(F 2
)
= (0.04) 2
= 0.0016
since events areindependent
(b) We requireboth components tobenot faulty:
P(F 1
∩F 2
)=P(F 1
)P(F 2
)
= (0.96) 2
= 0.9216
(c) Consider the two components. Then either both components are faulty or one
component isfaulty or neither of the components isfaulty. So,
P(one component isfaulty) = 1 −P(both components are faulty)
−P(neither of the components isfaulty)
= 1 −0.0016 −0.9216
= 0.0768
Analternativeapproachisasfollows.Eitherthefirstcomponentisfaultyandthe
second one is not faulty, or the first component is not faulty and the second one
isfaulty. These two cases arerepresented byF 1
∩F 2
andF 1
∩F 2
.So,
P(one component isfaulty) =P(F 1
∩F 2
)+P(F 1
∩F 2
)
=P(F 1
)P(F 2
)+P(F 1
)P(F 2
)
= (0.04)(0.96) + (0.96)(0.04)
= 0.0768
(d) Werequireonecomponenttobenotfaulty.Sincetherearetwocomponentsthen
requiring one to be not faulty is equivalent to requiring one to be faulty. Hence
the calculation isthe same as thatin(c):
P(one component isnot faulty) = 0.0768
(e) At least one of the components is faulty means that one or both of the components
isfaulty:
➔