Answer to Question #345892 in Statistics and Probability for Toilet Gamer

Question

Q: Suppose a dataset has 8500 email collection. Among 8500 emails, 4000 emails are not-spam and remaining are spam emails. The word “dating” is used as a feature, whose frequency/count in spam emails are 310 and 106 in not-spam emails. You have to compute two probabilities using bayes theorem, only knowing it contains the word “dating”.

 

First: Probability of an email being spam?        Second: Probability of an email being not spam?

Answer 1

Let $S$ denote the event "spam email".Let $D$ denote the event "dating".

Given

$$P(S^C)=\dfrac{4000}{8500}=\dfrac{8}{17}, P(S)=\dfrac{9}{17}.$$

$$P(D| S)=\dfrac{106}{106+310}=\dfrac{53}{208}, P(D| S^C)=\dfrac{155}{208}$$

a)

$$P(S|D)=\dfrac{P(D|S)P(S)}{P(D|S)P(S)+P(D|S^C)P(S^C)}$$

$$=\dfrac{\dfrac{53}{208}(\dfrac{9}{17})}{\dfrac{53}{208}(\dfrac{9}{17})+\dfrac{155}{208}(\dfrac{8}{17})}=0.2778$$

b)

$$P(S^C|D)=\dfrac{P(D|S^C)P(S^C)}{P(D|S)P(S)+P(D|S^C)P(S^C)}$$

$$=\dfrac{\dfrac{155}{208}(\dfrac{8}{17})}{\dfrac{53}{208}(\dfrac{9}{17})+\dfrac{155}{208}(\dfrac{8}{17})}=0.7222$$