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

904 Chapter 28 Probability
E
A
28.2 INTRODUCINGPROBABILITY
B = A
Figure28.1
A:acomponent meets
the specification.B =A:
acomponent fails to
meet the specification.
Consideramachinewhichmanufactureselectroniccomponents.Thesemustmeetacertain
specification. The quality control department regularly samples the components.
Suppose, on average, 92 out of 100 components meet the specification. Imagine that a
component is selected at random and letAbe the outcome that a component meets the
specification; let B be the outcome that a component does not meet the specification.
Then we say the probability of A occurring is 92 = 0.92 and the probability of B
100
8
occurring is = 0.08. The probability is thus a measure of the likelihood of the
100
occurrence of a particular outcome. We write
P(A) = probability ofAoccurring = 0.92
P(B) = probability ofBoccurring = 0.08
We note that the sum of the probabilities of all possible outcomes is 1. The process of
selecting a component is called a trial. The possible outcomes are also called events.
In this example there are only two possible events,AandB. We can depict this situation
using a Venn diagram as shown in Figure 28.1. Recall from Section 5.2 that Venn
diagrams areusedtodepictsets.InthisdiagramwearedepictingtheeventsAandBas
sets. The set of all possible outcomes is called the sample space and is represented by
the universal set E. The setArepresents the event that a component meets the specification.
The setBrepresents the event that a component fails to meet the specification.
In this instance, when a trial takes place there are only two possible outcomes, either a
component meets the specification or it does not; that is, either eventAoccurs or event
Boccurs. An alternative notation istowrite
B =A
whereAissaidtobethecomplementofA. We could alsowrite
A =B
When a bar appears over a set then wesay, for example, ‘notA’, or‘notB’.
Wenowseektodefineprobabilityinamoreformalway.LetE beanevent.Wewish
toobtaintheprobabilitythatE willoccur,thatisP(E),whenatrialtakesplace.Inorder
to do so we repeat the trial a large number of times,n. We count the number of times
that eventE occurs,denoted bym. We then conclude that
P(E)= m n
(28.1)
Thelargerthenumberoftrialsthattakeplace,themoreconfidentweareofourestimate
of the probability of E occurring. For example, consider the trial of tossing a coin. If
we wish to calculate the probability of a head occurring then measuring the results of
1000 tosses of the coin is likely to yield a more accurate estimate than measuring the
results of 10 tosses of the coin. Various consequences flow from Equation (28.1). The
numberoftimesE occursmustbenon-negativeandlessthanorequaltothenumberof
trials,thatis0 mn. So,
0P(E)1