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A First Course in Probability

Sheldon M. Ross

$Math Studyset 91影视 Explanations$ Math

9th Edition 路 432 pages

ISBN: 9780321794772

Chapter 2: Q. 2.14 (page 55)

Prove Boole鈥檚 inequality:
$P ({鈭�}_{i = 1}^{鈭�} A_{i}) 鈮� {鈭�}_{i = 1}^{鈭�} P (A_{i})$

Short Answer

Expert verified

Proof by mathematical induction:

Assume that equality stands for some $n$ , and it follows that inequality stands for $n + 1$ .

Step by step solution

01

Given Information.

Prove that for all events $A_{1}, A_{2}, . . . A_{n}$ .

$P ({鈭�}_{k = 1}^{n} A_{k}) 鈮� {鈭�}_{k = 1}^{n} P (A_{k})$

02

Explanation.

Proof by mathematical induction:

For $n = 1$ the statement is true

$P (A_{1}) 鈮� P (A_{1})$

Hypothesis: $P ({鈭�}_{k = 1}^{n} A_{k}) 鈮� {鈭�}_{k = 1}^{n} P (A_{k})$ this is true for any $n$ event.

Then for any $n + 1$ events:

$P ({鈭�}_{k = 1}^{n + 1} A_{k}) = P [({鈭�}_{k = 1}^{n} A_{k}) 鈭� A_{n + 1}] associative law$

& $= P ({鈭�}_{k = 1}^{n} A_{k}) + P (A_{n + 1}) - P [({鈭�}_{k = 1}^{n} A_{k}) 鈭� A_{n + 1}] 鈮� {鈭�}_{k = 1}^{n} P (A_{k}) + P (A_{n + 1}) - P [({鈭�}_{k = 1}^{n} A_{k}) 鈭� A_{n + 1}] 鈮� {鈭�}_{k = 1}^{n + 1} P (A_{k}) - P [({鈭�}_{k = 1}^{n} A_{k}) 鈭� A_{n + 1}] 鈮� {鈭�}_{k = 1}^{n + 1} P (A_{k})$

$P (\overset{n + 1}{鈭�} A_{k}) 鈮� {鈭�}_{k = 1}^{n + 1} P (A_{k})$

This proves the statement, by the principle of mathematical induction.

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