Based on Bayes' theorem calculate mixed probability

Solution 1:

I'm gonna address the picture you quoted and not the tree diagram on the whiteboard.

There are six words ordered as seen in the table on the right: {password, review, send, us, your, account}.

If a word is in the sentence under examination, then it's a $1$, otherwise it's a $0$. This way we can encode the sentence as a string of binaries of $0$s and $1$s.

So we have a new email with the sentence "review us now". The last word "now" is ignored because it is not in the dictionary.

• The word password is not in it so it's a $0$.

• The word review is in so it's a $1$.

• send is not in so $0$.

• us is in so $1$

• your is not in so $0$

• account is not in so $0$.

This becomes $P(\text{review us}~|~\text{spam} ) = P(\{0,1,0,1,0,0\}~|~\text{spam} )$

For ones we have $p_i$ and for the zeros we calculate as $1 - p_i$, where $p_i$ is the corresponding conditional probability listed in the "spam" column in the same table: $$\{p_1,\, p_2,\, p_3,\, p_4,\, p_5,\, p_6\} = \{ \frac24, \, \frac14, \, \frac34, \, \frac34, \, \frac34, \, \frac14\}~.$$

The first word in the dictionary (first row) password is not in the sentence (encoded as $0$), and it has a conditional $p_1 = \frac24$. The contribution of this word to the likelihood is $1 - p_1 = (1 - \frac24)$

The second word in the dictionary (second row) review is in (thus encoded as $1$) and has a conditional $p_2 = \frac14$. Its contribution to the likelihood is $p_2 = (\frac14)$ where the parentheses are just to indicate the contributions from different words.

The third word in the dictionary ($3$rd row) send is NOT in, encoded as $0$, with a conditional $p_3 = \frac34$, and its contribution to the likelihood is $1 - p_3 = (1 - \frac34)$

So on and so forth.

The same procedure goes for $P(\text{review us}~|~\text{ham} ) = P(\{0,1,0,1,0,0\}~|~\text{ham} )$, with the same encoding (same dictionary) but calculated using the conditional probabilities of the other column (given that it is ham): $\{ \frac12, \, \frac22, \, \frac12, \, \frac12, \, \frac12, \, \frac02 \}$