3
$\begingroup$

enter image description here

I got that $P(A_1) = 0.07875, P(A_2) = 0.008, P(A_3) = 0.0625$, we know that

$$P(\text{Spam} \mid A_1A_2A_3) = \frac{P(\text{Spam} \cap A_1A_2A_3)}{P(A_1A_2A_3)}$$

We know that $\{A_1, A_2, A_3\}$ is an independent set., so $P(A_1A_2A_3) = 0.00039375$

I cant figure out $P(\text{spam} \times A_1A_2A_3)$?

But I'm not sure of numerator?

  • 0
    $$P(\mathrm{spam}\cap A_1\cap A_2\cap A_3)=P(\mathrm{spam})P(A_1\cap A_2\cap A_3\mid\mathrm{spam})=P(\mathrm{spam})P(A_1\mid\mathrm{spam})P(A_2\mid\mathrm{spam})P(A_3\mid\mathrm{spam})$$ – 2017-02-01
  • 0
    Note that you take for granted that $(A_k)$ is independent while the hypothesis is that $(A_k)$ is independent conditionally on spam and that $(A_k)$ is independent conditionally on non-spam. Can you show these indeed imply that $(A_k)$ is independent? – 2017-02-01

0 Answers 0