Need clarification on Naive Bayes Theorem

Course Q&A Math for Machine Learning Probability & Statistics for Machine Learning &...
2

Hi @Sandhya5

In this context, A corresponds to the event that the sum is 10, and B corresponds to the event that the first dice shows 6. So, we have:

  • Prior: P(A) = \frac{3}{36} = \frac{1}{12}
    -> (6,4), (4,6), (5,5) / All Combinations

  • Observed Event: P(B) = \frac{6}{36} = \frac{1}{6}
    -> (6,1), (6,2), (6,3), (6,4), (6,5), (6,6) / All Combinations

  • Likelihood: P(B | A) = \frac{P(A \cap B)}{P(A)} = \frac{1}{3}
    -> (6,4) / (6,4), (4,6), (5,5)

  • Posterior: P(A | B) = \frac{P(B | A) \cdot P(A)}{P(B)} = \frac{\frac{1}{3}\cdot \frac{1}{12}}{\frac{1}{6}} = \frac{6}{36} = \frac{1}{6}

Hope this helps. Feel free to ask if you need further assistance!

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