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The probabilities of states of independent systems X and Y are X1=1 X2=-2 Yj Y1=-1 Y2=0 0.4 0.3 0.7 Rj 0.6 System Z is related to systems X and Y by the relation Z = X + (X-Y)*2. Find complete information I(Z, Y) about system Y contained in system Z. given by tables: Xi Pi

The probabilities of states of independent systems X and Y are X1=1 X2=-2 Yj Y1=-1 Y2=0 0.4 0.3 0.7 Rj 0.6 System Z is related to systems X and Y by the relation Z = X + (X-Y)*2. Find complete information I(Z, Y) about system Y contained in system Z. given by tables: Xi Pi

A First Course in Probability (10th Edition)
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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The probabilities of states of independent systems X and Y are X1=1 X2=-2 Yj Y1=-1 Y2=0 0.4 0.3 0.7 Rj 0.6 System Z is related to systems X and Y by the relation Z = X + (X-Y)*2. Find complete information I(Z, Y) about system Y contained in system Z. given by tables: Xi Pi
Transcribed Image Text:The probabilities of states of independent systems X and Y are X1=1 X2=-2 Yj Y1=-1 Y2=0 0.4 0.3 0.7 Rj 0.6 System Z is related to systems X and Y by the relation Z = X + (X-Y)*2. Find complete information I(Z, Y) about system Y contained in system Z. given by tables: Xi Pi
Example 16. The probabilities of states of independent systems X and Y are given by tables: x₁|₁=0x₂=1 0,4 3516-1202 0,8 Pi 0,6 System Z is related to systems X and Y by the relation Z = X-(Y-X)². Find complete information I(Z. Y) about system Y contained in system Z. Solution. Let us find the states of the system Z. If x, = 0 and y₁ = -1, then z = -1. If x₁ = 0 and y₂ = 1, then z=-1. If x₂ = 1 and y₁ = -1, then z = -3. If x₂ = 1 and y₂ = 1, then z = 1. Let us find the probabilities of the states of the system Z: 2₁=-3 2₂=-1 23 = 1 qk0,6-0,8 0,48 0,4-0,8+0,4-0,2 = 0,4 0,6-0,2=0, 12 Let us find the probabilities of the states of the combined system (Y,Z). (Mj, 2k) 21 22 23 9/1 0,48 0,32 0 0 0,08 0,12 H(Y) = n(0,8)+7(0, 2) = 0, 2575 +0,4644 = 0,7219 H(Z) = n(0,48)+n(0, 4)+7(0, 12) = 0, 5083+0,5288+0,3671 = 1, 4042 H(Y,Z) = n(0,48) + n(0, 32)+n(0,08)+7(0, 12) = 0, 5083 +0,5260+ 0, 2915+0,3671=1,6929 I(Z,Y)= H(Z) + H(Y) - H(Y,Z) = 0, 4332 GT.
Transcribed Image Text:Example 16. The probabilities of states of independent systems X and Y are given by tables: x₁|₁=0x₂=1 0,4 3516-1202 0,8 Pi 0,6 System Z is related to systems X and Y by the relation Z = X-(Y-X)². Find complete information I(Z. Y) about system Y contained in system Z. Solution. Let us find the states of the system Z. If x, = 0 and y₁ = -1, then z = -1. If x₁ = 0 and y₂ = 1, then z=-1. If x₂ = 1 and y₁ = -1, then z = -3. If x₂ = 1 and y₂ = 1, then z = 1. Let us find the probabilities of the states of the system Z: 2₁=-3 2₂=-1 23 = 1 qk0,6-0,8 0,48 0,4-0,8+0,4-0,2 = 0,4 0,6-0,2=0, 12 Let us find the probabilities of the states of the combined system (Y,Z). (Mj, 2k) 21 22 23 9/1 0,48 0,32 0 0 0,08 0,12 H(Y) = n(0,8)+7(0, 2) = 0, 2575 +0,4644 = 0,7219 H(Z) = n(0,48)+n(0, 4)+7(0, 12) = 0, 5083+0,5288+0,3671 = 1, 4042 H(Y,Z) = n(0,48) + n(0, 32)+n(0,08)+7(0, 12) = 0, 5083 +0,5260+ 0, 2915+0,3671=1,6929 I(Z,Y)= H(Z) + H(Y) - H(Y,Z) = 0, 4332 GT.
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