Probability & Statistics with R for Engineers and Scientists
Probability & Statistics with R for Engineers and Scientists
1st Edition
ISBN: 9780321852991
Author: Michael Akritas
Publisher: PEARSON
Question
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Chapter 2.6, Problem 8E
To determine

Show that events E1,E2,E3 are pairwise independent.

Expert Solution & Answer
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Explanation of Solution

Calculation:

Probability:

The formula for probability of event E is,

P(E)=N(E)N

In the formula, N(E) is the number of outcomes for event E and N is the total number of outcomes.

Pairwise independence:

Let E1, E2, E3 be three events, then events are said to be pairwise independent if,

P(E1E2)=P(E1)P(E2)P(E1E3)=P(E1)P(E3)P(E2E3)=P(E2)P(E3)

Also, pairwise independence does not imply P(E1E2E3)=P(E1)P(E2)P(E3).

The event E1 is defined as sum of the two outcomes is 7 and event E2 is defined as outcome of the first roll is 3 and event E3 is defined as outcome of the second roll is 4.

A die has 6 outcomes {1,2,3,4,5,6}. When a die is rolled twice the outcomes are,

S={(1,1),(1,2),(1,3),(1,4),(1,5),(1,6)(2,1),(2,2),(2,3),(2,4),(2,5),(2,6)(3,1),(3,2),(3,3),(3,4),(3,5),(3,6)(4,1),(4,2),(4,3),(4,4),(4,5),(4,6)(5,1),(5,2),(5,3),(5,4),(5,5),(5,6)(6,1),(6,2),(6,3),(6,4),(6,5),(6,6)}

The total possible outcomes are, N=36.

The favorable outcomes for event E1 are {(1,6),(2,5),(3,4),(4,3),(5,2),(6,1)}. The probability is,

P(E1)=636=0.1667

The favorable outcomes for event E2 are {(3,1),(3,2),(3,3),(3,4),(3,5),(3,6)}. The probability is,

P(E2)=636=0.1667

The favorable outcomes for event E3 are {(1,4),(2,4),(3,4),(4,4),(5,4),(6,4)}. The probability is,

P(E3)=636=0.1667

The favorable outcome for event E1E2 is (3,4). The probability is,

P(E1E2)=136=0.0278

The favorable outcome for event E1E3 is (3,4). The probability is,

P(E1E3)=136=0.0278

The favorable outcome for event E2E3 is (3,4). The probability is,

P(E2E3)=136=0.0278

Independence of events E1 and E2:

P(E1)P(E2)=0.1667×0.1667=0.0278=P(E1E2)

Independence of events E1 and E3:

P(E1)P(E3)=0.1667×0.1667=0.0278=P(E1E3)

Independence of events E2 and E3:

P(E2)P(E3)=0.1667×0.1667=0.0278=P(E2E3)

It can be observed that, the events E1,E2,E3 are pairwise independent.

The favorable outcome for E1E2E3 is (3,4). The probability is,

P(E1E2E3)=136=0.0278

Independence of events E1, E2 and E3:

P(E1)P(E2)P(E3)=0.1667×0.1667×0.1667=0.0046P(E1E2E3)

Hence, the pairwise independence does not imply P(E1E2E3)=P(E1)P(E2)P(E3).

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