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

926 Chapter 28 Probability
P(E 2
|E 1
) is the probability ofE 2
happening givenE 1
has happened. However, machine1andmachine2areindependent,sotheprobabilityofchipBbeingacceptable
is in no way influenced by the acceptability of chip A. The events E 1
and E 2
are
independent.
Therefore
P(E 2
|E 1
) =P(E 2
) = 0.83
P(E 1
∩E 2
) =P(E 1
)P(E 2
) = (0.9)(0.83) = 0.75
For independent eventsE 1
andE 2
:
(1)P(E 1
|E 2
) =P(E 1
), P(E 2
|E 1
) =P(E 2
)
(2)P(E 1
∩E 2
) =P(E 1
)P(E 2
)
The concept of independence may be applied to more than two events. Three or more
events are independent if every pair of events is independent. If E 1
,E 2
,...,E n
are n
independent events then
P(E i
|E j
)=P(E i
)
foranyiandj,i≠j
and
P(E i
∩E j
) =P(E i
)P(E j
)
i≠j
Acompoundeventmaycompriseseveralindependentevents.Themultiplicationlawis
extended inanobvious way:
P(E 1
∩E 2
∩E 3
) =P(E 1
)P(E 2
)P(E 3
)
P(E 1
∩E 2
∩E 3
∩E 4
) =P(E 1
)P(E 2
)P(E 3
)P(E 4
)
and soon.
Engineeringapplication28.15
Probabilityoffaultycomponents
The probability that a component is faulty is 0.04. Two components are picked at
random. Calculate the probability that
(a) both components arefaulty
(b) both components arenot faulty
(c) one ofthe components isfaulty
(d) one ofthe components isnotfaulty
(e) atleastone of the components isfaulty
(f) atleastone of the components isnot faulty