Doubt regarding the naive bayes algorithm

Why is P(lottery & winning | spam) is equivalent to the probability of intersection between two events(P(A \cap B) as shown in the video)? I am completely stuck at this point. Please elaborate on this help me build an intuition.

Hi @Rhythm_Dutta!

The intersection is only between lottery and winning. To be very consistent, it should be P(A \cap B \mid \text{spam}). However, to avoid excess of notation and to keep the main idea, it is shown only P(A \cap B).

Cheers,
Lucas

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Okay! That means we CAN write:
P(A \cap B \mid \text{spam}) = P(A \mid \text{spam}) \times P(B \mid \text{spam}) for independent events A and B?

You are correct.

This is a bit more deep than we covered in the course, but if we look to the conditional probability P( \cdot \mid spam), it defines a probability, therefore every probability rule will also hold for the conditional probability too.

Lucas

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