Assume that P(A₁ N A₂ NA3) > 0. Prove that P(A₁ NA₂ NÃ3 Ñ A4) = P(A1)P(A2|A1₁)P(A3|A1 N A₂)P(A4|A₁ ÑA₂ Ñ A3)

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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Assume that P(A₁ N A₂ Ñ A3) > 0. Prove that
P(A1 A₂ A3 n A4) = P(A₁)P(A₂|A₁)P(A3|A₁ MA₂)P(A4|A₁ A₂ A3)
Transcribed Image Text:Assume that P(A₁ N A₂ Ñ A3) > 0. Prove that P(A1 A₂ A3 n A4) = P(A₁)P(A₂|A₁)P(A3|A₁ MA₂)P(A4|A₁ A₂ A3)
Expert Solution
Step 1: Conditional probabilities obtained.

We have conditional probabilities

P(A subscript 4 vertical line A subscript 1 intersection A subscript 2 intersection A subscript 3) = fraction numerator P left parenthesis A subscript 1 intersection A subscript 2 intersection A subscript 3 intersection A subscript 4 right parenthesis over denominator P left parenthesis A subscript 1 intersection A subscript 2 intersection A subscript 3 right parenthesis end fraction      (i)

P(A subscript 3 vertical line A subscript 1 intersection A subscript 2 right parenthesisfraction numerator P left parenthesis A subscript 1 intersection A subscript 2 intersection A subscript 3 right parenthesis over denominator P left parenthesis A subscript 1 intersection A subscript 2 right parenthesis end fraction                      (ii)

P(A subscript 2 vertical line A subscript 1 right parenthesis =fraction numerator P left parenthesis A subscript 1 intersection A subscript 2 right parenthesis over denominator P left parenthesis A subscript 1 right parenthesis end fraction                                       (iii)

Therefore, in view of equations (i),(ii) and (iii),we have

P(A subscript 1 right parenthesis space P left parenthesis A subscript 2 vertical line A subscript 1 right parenthesis space P left parenthesis A subscript 3 vertical line A subscript 1 intersection A 2 right parenthesis space P left parenthesis A subscript 4 vertical line A subscript 1 intersection A subscript 2 intersection A subscript 3 right parenthesis =P(A subscript 1 intersection A subscript 2 intersection A subscript 3 intersection A subscript 4 right parenthesis

Hence proved.

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