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Get Full Access to Probability And Statistical Inference - 9 Edition - Chapter 1.4 - Problem 4e
Get Full Access to Probability And Statistical Inference - 9 Edition - Chapter 1.4 - Problem 4e

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# Prove parts (b) and (c) of Theorem 1.4-1.

ISBN: 9780321923271 41

## Solution for problem 4E Chapter 1.4

Probability and Statistical Inference | 9th Edition

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Problem 4E

Prove parts (b) and (c) of Theorem 1.4-1.

Step-by-Step Solution:
Step 1 of 3

Solution 4E

Step1 of 3:

We have the events A and B are independent events.

We need to prove,

a).A1 and B are independent.

b).A1 and B1  are independent.

Step2 of 3:

a).

Let us assume that the events A and B are independent events

Consider,

P(A1 B) = P(A1/ B)P(B)                 [because P(A/ B) = ]

= [1 - P(A/B)]P(B)

= P(B) - P(A/B)P(B)

= P(B) - P(A B)                [because P(A/ B) = ]

= P(B) - [P(A)P(B)]         [because we know that A and B are independent]

= P(B)[1 - P(A)]

= P(B) P(A1).

Therefore, P(A1 B) = P(A1)P(B)

Hence the proof.

Step3 of 3:

b).

Let us assume that the events A and B are independent events

Consider,

P(A1 B1) = [P(AB)1]

= [1 - P(AB)]

= {1 - [P(A) + P(B) - P(AB)]}

= {1 - P(A) - P(B) + P(AB)}

= P(A1) - P(B) + P(AB)                 [because P(A1) = 1 - P(A)]

= P(A1) - P(B) + P(A)P(B)     [because we know that A and B are independent]

= P(A1) - P(B)[1 - P(A)]             [because P(A1) = 1 - P(A)]

= P(A1) - P(B)P(A1)

= P(A1) [1 - P(B)]

= P(A1)P(B1)                                                    [because P(B1) = 1 - P(B)]

Therefore, P(A1 B1) = P(A1)P(B1)

Hence the proof.

Step 2 of 3

Step 3 of 3

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