Supply the missing quantities. DescriptionA tree diagram has two stages. Stage 1 has two nodes and stage 2 has four nodes.In stage 1, the branch from the starting position to node A is labeled 0.6. The branch from the starting position to node B is an answer blank.In stage 2, the branch from node A to node C is labeled 0.3. The branch from node A to node D is an answer blank.In stage 2, the branch from node B to node C is an answer blank. The branch from node B to node D is labeled 0.7.OutcomeP(A ∩ C)=P(A ∩ D)=P(B ∩ C)=P(B ∩ D)=

P(A∩C) = 0.12 P(A∩D) = 0.18 P(B∩C) = 0.28 P(B∩D) = 0.42P(A∩C) =0.3*0.4= 0.12 P(A∩D) =0.3*0.6= 0.18 P(B∩C) =0.7*0.4= 0.28 P(B∩D) =0.7*0.6= 0.42 Answer 2: Now, P(A)=0.4 , P(B)=0.1 , A and B are independent P(A∩B) = 0.4*0.1= 0.04 Answer 3: Probability = 0.74*0.36 = 0.27 Hope you understand the logic.

Answered: Description A tree diagram has two…

Database System Concepts

7th Edition

ISBN:9780078022159

Author:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan

Publisher:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan

Chapter1: Introduction

Section: Chapter Questions

Problem 1PE

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Question

Supply the missing quantities.

Description

A tree diagram has two stages. Stage 1 has two nodes and stage 2 has four nodes.

In stage 1, the branch from the starting position to node A is labeled 0.6. The branch from the starting position to node B is an answer blank.
In stage 2, the branch from node A to node C is labeled 0.3. The branch from node A to node D is an answer blank.
In stage 2, the branch from node B to node C is an answer blank. The branch from node B to node D is labeled 0.7.

Outcome

P(A ∩ C)

P(A ∩ D)

P(B ∩ C)

P(B ∩ D)

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