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Unformatted text preview: = P (no counts in (0,1]) · P (two counts in (1,2]) · P (no counts in (2,3])
λ2 e−λ
λ2 −3λ
= e−λ
e−λ =
e.
2!
2
So, by the law of total probability,
P (A) = P (B020 ) + P (B111 ) + P (B202 )
λ2 −3λ
λ2 e−λ
=
e
+ (λe−λ )3 +
e−λ
2
2 λ2 e−λ
2 = λ2
λ4
+ λ3 +
2
4 e−3λ . (c) By the deﬁnition of conditional probability,
P (B020 A) =
= P (B020 A)
P (B020 )
=
P (A)
P (A)
λ2
2 + λ2
2
λ3 + λ4
4 = 2
.
2 + 4λ + λ2 Notice that in part (b) we applied the law of total probability to ﬁnd A, and in part (c) we applied
the deﬁnition of the conditional probability P (B020 A). Together, this amounts to application of
Bayes rule for ﬁnding P (B020 A). 86 3.5.3 CHAPTER 3. CONTINUOUSTYPE RANDOM VARIABLES The gamma distribution Let Tr denote the time of the rth count of a Poisson process. Thus, Tr = U1 + · · · + Ur , where
U1 , . . . , Ur are independent, exponentially distributed random variables with parameter λ. This
characterization of Tr and the method of Section 4.5.2, showing how to ﬁnd the pdf of the sum of
independent continuoustype random vari...
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This note was uploaded on 02/09/2014 for the course ISYE 2027 taught by Professor Zahrn during the Spring '08 term at Georgia Institute of Technology.
 Spring '08
 Zahrn
 The Land

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