Naive bayes problem

Question

I have this problem:

You have two hypotheses: 
H1 ="the weather tomorrow will be good" and 
H2 ="the weather tomorrow will be bad". 
Let D1 be the observation that the sky is red in the evening. 
Let D2 be the observation that weather.gov forecasts tomorrow will be good.

I am also given: Over the last last 1000 days, there have been 200 good days and 800 bad days. 160 of the good days were forecast to be good and 200 of the bad days were forecast to be good. 100 of the good days were preceded by a red sky at night and 300 of the bad days were preceded by a red sky at night. What values would you pick for the priors

P(H1); P(H2) and the likelihoods P(D1|H1); P(D2|H1); P(D1|H2); P(D2|H2)?

So I have found out these calculations:

P(H1) = 200/1000
P(H2) = 800/1000
P(D1|H1) = 100/200
P(D2|H1) = 160/200
P(D1|H2) = 300/800
P(D2|H2) = 200/800 

Now I have to answer the following:

Suppose you observe that the sky is red and the weather.gov forecasts good weather tomorrow.Using a Naive Bayes Classier, would you deduce that the weather will be good tomorrow?

How exactly do I go about doing this? I am very stuck with this. I know the formula but I am not sure exactly how to apply it here.

Any help is useful. Thank you.

Answer 1

Calculate the posterior numerators for $H_1, H_2$ using your prior probabilities:

\begin{eqnarray*} p_n(H_1) &=& P(H_1)P(D_1\mid H_1)P(D_2\mid H_1) \\ &=& \frac{1}{5} \times \frac{1}{2} \times \frac{4}{5} = \frac{2}{25} = 0.08 \\ \end{eqnarray*}

\begin{eqnarray*} p_n(H_2) &=& P(H_2)P(D_1\mid H_2)P(D_2\mid H_2) \\ &=& \frac{4}{5} \times \frac{3}{8} \times \frac{1}{4} = \frac{3}{40} = 0.075 \\ \end{eqnarray*}

We have $p_n(H_1) \gt p_n(H_2)$ so we deduce that the weather will be good tomorrow.