Question

2. Giving the following training dataset: B Label Image 1 0-100 171-255 0-150 Yes Tree Image 2 101-170 0-100 151-255 No not Tree Image 3 171-255 101-170 151-255 Yes Tree Image 4 101-170 0-100 151-255 No not Tree Image 5 171-255 101-170 151-255 No Tree Image 6 101-170 0-100 0-150 No not Tree Image 7 101-170 0-150 Yes not Tree 0-100 0-100 Image 8 101-170 151-255 Yes Tree Image 9 101-170 171-255 0-150 No not Tree Image 10 171-255 101-170 151-255 Yes not Tree Use Naive Bayes to predict to label of the image either Tree or not Tree using the following testing set B 3D Label Image A 0-100 0-100 0-150 Yes Image B 171-255 171-255 0-150 No Image C 171-255 101-170 151-255 Yes You will need to show Naive Bayes model and how use it to decide the outcome.

Answer 1

In the given training dataset:

Number of images with (Label=Tree) = 4

Number of images with (Label=not Tree) = 6

Probability of value of R being 0-100 when (Label=Tree) is given = P(R=0-100 | Label=Tree) = $\frac{Number\hspace{1mm}of\hspace{1mm}images\hspace{1mm}with\hspace{1mm}R=0-100\hspace{1mm}which\hspace{1mm}have\hspace{1mm}Label=Tree }{Number\hspace{1mm}of\hspace{1mm}images\hspace{1mm}with\hspace{1mm}Label=Tree}$ =2/4 = 1/2

Probability of value of R being 0-100 when (Label=not Tree) is given = P(R=0-100 | Label = not Tree) = $\frac{Number\hspace{1mm}of\hspace{1mm}images\hspace{1mm}with\hspace{1mm}R=0-100\hspace{1mm}which\hspace{1mm}have\hspace{1mm}Label=not\hspace{1mm}Tree }{Number\hspace{1mm}of\hspace{1mm}images\hspace{1mm}with\hspace{1mm}Label=not\hspace{1mm}Tree}$ =1/6

Similarly,

P(R=171-255 | Label=Tree) = 2/4 = 1/2

P(R=171-255 | Label=not Tree) = 1/6 = 1/6

P(G=0-100 | Label=Tree) = 0/4 = 0

P(G=0-100 | Label=not Tree) = 3/6 = 1/2

P(G=171-255 | Label=Tree) = 1/4

P(G=171-255 | Label=not Tree) = 1/6

P(G=101-170 | Label=Tree) = 3/4

P(G=101-170 | Label=not Tree) = 2/6 = 1/3

P(B=0-150 | Label=Tree) = 1/4

P(B=0-150 | Label=not Tree) = 3/6 = 1/2

P(B=151-255 | Label=Tree) = 3/4

P(B=151-255 | Label=not Tree) = 3/6 = 1/2

P(3D=Yes | Label=Tree) = 3/4

P(3D=Yes | Label=not Tree) = 2/6 = 1/3

P(3D=No | Label=Tree) = 1/4

P(3D=No | Label=not Tree) = 4/6 = 2/3

For the testing set:

1) For Image A,

P(Image A | Label=Tree) = P(R=0-100 | Label=Tree)*P(G=0-100 | Label=Tree)*P(B=0-150 | Label=Tree)*P(3D=Yes | Label=Tree) = (1/2)*0*(1/4)*(3/4) = 0

P(Image A | Label=not Tree) = P(R=0-100 | Label=not Tree)*P(G=0-100 | Label=not Tree)*P(B=0-150 | Label=not Tree)*P(3D=Yes | Label=not Tree) = (1/6)*(1/2)*(1/2)*(1/3) = 1/72

According to Naive Bayes classifier, Label's output will be according to whichever one's value is higher. We can see that (1/72) > 0. So, Label = not Tree for Image A.

2) For Image B,

P(Image B | Label=Tree) = P(R=171-255 | Label=Tree)*P(G=171-255 | Label=Tree)*P(B=0-150 | Label=Tree)*P(3D=No | Label=Tree) = (1/2)*(1/4)*(1/4)*(1/4) = 1/128

P(Image B | Label=not Tree) = P(R=171-255 | Label=not Tree)*P(G=171-255 | Label=not Tree)*P(B=0-150 | Label=not Tree)*P(3D=No | Label=not Tree) = (1/6)*(1/6)*(1/2)*(2/3) = 1/108

According to Naive Bayes classifier, Label's output will be according to whichever one's value is higher. We can see that (1/108) > (1/128). So, Label = not Tree for Image B.

3) For Image C,

P(Image C | Label=Tree) = P(R=171-255 | Label=Tree)*P(G=101-170 | Label=Tree)*P(B=151-255 | Label=Tree)*P(3D=Yes | Label=Tree) = (1/2)*(3/4)*(3/4)*(3/4) = 27/128 = 0.2109

P(Image C | Label=not Tree) = P(R=171-255 | Label=not Tree)*P(G=101-170 | Label=not Tree)*P(B=151-255 | Label=not Tree)*P(3D=Yes | Label=not Tree) = (1/6)*(1/3)*(1/2)*(1/3) = 1/108 = 0.009

According to Naive Bayes classifier, Label's output will be according to whichever one's value is higher. We can see that (0.2109) > (0.009). So, Label = Tree for Image C.