A QR code photographed in poor lighting, so that it can be difficult to distinguish black and white pixels. The gray color (X) in each pixel is therefore coded on a scale from 0 (white) to 100 (black). The true pixel value (without shadow) the code is Y = 0 for white, and Y = 1 for black. We treat X and Y as random variables. For the highlighted pixel in the figure is the gray color X = 20 and the true pixel value is white, i.e. Y = 0. We assume that QR codes are designed so that, on average, there are as many white as black pixels, which means that pY (0) = pY (1) = 1/2. In this situation, X is continuously distributed (0 ≤ X ≤ 100) and Y is discretely distributed, but we can still think about th

Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter4: Polynomial And Rational Functions
Section4.5: Rational Functions
Problem 48E
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A QR code photographed in poor lighting, so that it can be difficult to distinguish black and white pixels. The gray color (X) in each pixel is therefore coded on a scale from 0 (white) to 100 (black). The true pixel value (without shadow) the code is Y = 0 for white, and Y = 1 for black. We treat X and Y as random variables. For the highlighted pixel in the figure is the gray color X = 20 and the true pixel value is white, i.e. Y = 0. We assume that QR codes are designed so that, on average, there are as many white as black pixels, which means that pY (0) = pY (1) = 1/2. In this situation, X is continuously distributed (0 ≤ X ≤ 100) and Y is discretely distributed, but we can still think about the simultaneous distribution of X and Y. We start by defining the conditional density of X, given the value of Y : 
fX|Y(x|0) = "Pixel is really white"

fX|Y(x|1) =" Pixel is really balck "

Use Bayes formula as given in the picture and find the probability for x = 20 like in the picture. 

| fx(2) =Zfx\x(\)pY(g) 0 ≤x≤100.
x 2
100
2
3
fx|y(x0)
fxy(x1) God
100 100.
|
100
3
(1 –
P(Y=0|X = x) =
fxy(x0)py (0)
fx(x)
Transcribed Image Text:| fx(2) =Zfx\x(\)pY(g) 0 ≤x≤100. x 2 100 2 3 fx|y(x0) fxy(x1) God 100 100. | 100 3 (1 – P(Y=0|X = x) = fxy(x0)py (0) fx(x)
25
ETS
ドロ
Transcribed Image Text:25 ETS ドロ
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