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From observations of the weather, it is known that on October 20 the probability of rain is 0.85. A certain forecast method for October 20 is correct  0.9 times if rain is predicted and 0.75 times if no rain is predicted.  How much information does a real weather forecast provide 20 th of October?

From observations of the weather, it is known that on October 20 the probability of rain is 0.85. A certain forecast method for October 20 is correct  0.9 times if rain is predicted and 0.75 times if no rain is predicted.  How much information does a real weather forecast provide 20 th of October?

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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From observations of the weather, it is known that on October 20 the probability of rain is 0.85. A certain forecast method for October 20 is correct  0.9 times if rain is predicted and 0.75 times if no rain is predicted.  How much information does a real weather forecast provide 20 th of October?

n.b the picture is example of solution

Task 14. The probability of a shooter hitting the target with one shot is 0.15. The shooter fires k independent shots at the target. After that, a message arrives whether the target is hit or not. If the target is hit, the shooting stops. At what k will the amount of information contained in the message be the largest? Solution. The physical system X is the state of the target after the k-th shot. System X has two states: x1 цель поражена P₁=1 (1-0, 15) The information contained in the target state message will be maximum if both states X₁ and x₂ equiprobable: p₁ = P2. 1- (1-0, 15) = (1-0, 15) log/ X2 цель не поражена P2 = (1-0, 15) ⇒2. (1-0, 15) = 1 ⇒log(1-0, 15)* = k = log(1=0,15) = 4 выстрела. k log(1-0, 15) = -1
Transcribed Image Text:Task 14. The probability of a shooter hitting the target with one shot is 0.15. The shooter fires k independent shots at the target. After that, a message arrives whether the target is hit or not. If the target is hit, the shooting stops. At what k will the amount of information contained in the message be the largest? Solution. The physical system X is the state of the target after the k-th shot. System X has two states: x1 цель поражена P₁=1 (1-0, 15) The information contained in the target state message will be maximum if both states X₁ and x₂ equiprobable: p₁ = P2. 1- (1-0, 15) = (1-0, 15) log/ X2 цель не поражена P2 = (1-0, 15) ⇒2. (1-0, 15) = 1 ⇒log(1-0, 15)* = k = log(1=0,15) = 4 выстрела. k log(1-0, 15) = -1
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