If the events B1, B2, ..., Bk constitute a partition of the sample space S such that P(B;) + 0 for i = 1, 2,. k, then for any event A of S, k P(A) = P(B; n A) = P(B÷)P(A|B;).
If the events B1, B2, ..., Bk constitute a partition of the sample space S such that P(B;) + 0 for i = 1, 2,. k, then for any event A of S, k P(A) = P(B; n A) = P(B÷)P(A|B;).
A First Course in Probability (10th Edition)
10th Edition
ISBN:9780134753119
Author:Sheldon Ross
Publisher:Sheldon Ross
Chapter1: Combinatorial Analysis
Section: Chapter Questions
Problem 1.1P: a. How many different 7-place license plates are possible if the first 2 places are for letters and...
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
Transcribed Image Text:If the events B1, B2, . ., Bk constitute a partition of the sample space S such that|
|P(B¡)+0 for i = 1, 2, . ., k, then for any event A of S,
P(B;) #0 for i = 1, 2, ...,
k, then for any event A of S,
k
k
P'A) = ΣPB; η Α) = Σ ΡΒ )P(Α | Β:) .
i=1
i=1
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B1
B5
BA
A
B2
Figure 14: Partitioning the sample space S."
Transcribed Image Text:Вз
B1
B5
BA
A
B2
Figure 14: Partitioning the sample space S.
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