Question 12 (1 point)

When calculating P(E | H), the Nave Bayes algorithm makes the assumption that

Question 12 options:

	the attributes are independent of each other
	the attributes are independent of the class
	the class is independent of the attributes
	attributes are all nominal
	all of the above

Question 13 (1 point)

Consider the Nave Bayes algorithm applied to the dataset in Table 1. What is P(Aquatic = yes | Class = mammal)? Assume a Laplace Estimator of 1 is added to the counts.

Table 1: Vertebrate Dataset (modified from Introduction to Data Mining by Tan et al.)

Name	Body temp	Give birth	Aquatic	Class
human	warm	yes	no	mammal
echidna	warm	no	no	mammal
salmon	cold	no	yes	fish
whale	warm	yes	yes	mammal
eel	cold	no	yes	fish
bat	warm	yes	no	mammal
shark	cold	yes	yes	fish
cat	warm	yes	no	mammal

Question 13 options:

	1/5
	2/6
	2/7
	4/5
	none of the above

Question 14 (1 point)

Consider the Nave Bayes algorithm applied to the dataset in Table 1. Assume a Laplace Estimator of 1 is added to the counts (not the class counts). What is the likelihood (not the probability) for the test instance in Table 2 being a mammal?

Table 1: Vertebrate Dataset (modified from Introduction to Data Mining by Tan et al.)

Name	Body temp	Give birth	Aquatic	Class
human	warm	yes	no	mammal
echidna	warm	no	no	mammal
salmon	cold	no	yes	fish
whale	warm	yes	yes	mammal
eel	cold	no	yes	fish
bat	warm	yes	no	mammal
shark	cold	yes	yes	fish
cat	warm	yes	no	mammal

Table 2: Test Instance

Name	Body temp	Give birth	Aquatic	Class
porcupine	warm	yes	no	?

Question 14 options:

	55454558{"version":"1.1","math":"$\frac{5}{5}\times \frac{4}{5}\times \frac{4}{5}\times \frac{5}{8}$"}
	67575758{"version":"1.1","math":"$\frac{6}{7}\times \frac{5}{7}\times \frac{5}{7}\times \frac{5}{8}$"}
	15251538{"version":"1.1","math":"$\frac{1}{5}\times \frac{2}{5}\times \frac{1}{5}\times \frac{3}{8}$"}
	67575658{"version":"1.1","math":"$\frac{6}{7}\times \frac{5}{7}\times \frac{5}{6}\times \frac{5}{8}$"}
	none of the above