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How do you verify Bayes' theorem using a set counting example with nine equally likely outcomes?

In a sample space of nine equally likely outcomes, let event A contain 5 elements and event B contain 6 elements, with their intersection containing 2 elements. The conditional probability P(A∣B)P(A \mid B) is calculated directly as the size of the intersection divided by the size of B: 2/6=1/32/6 = 1/3. Using Bayes' theorem, P(A)=5/9P(A) = 5/9, P(B)=6/9P(B) = 6/9, and P(B∣A)=2/5P(B \mid A) = 2/5. The formula gives P(A∣B)=(2/5)(5/9)6/9=2/96/9=2/6=1/3P(A \mid B) = \frac{(2/5)(5/9)}{6/9} = \frac{2/9}{6/9} = 2/6 = 1/3. Both methods yield the same result, verifying the theorem.

Conditions

  • The sample space has nine equally likely outcomes.
  • Event A has 5 elements, Event B has 6 elements, and A∩BA \cap B has 2 elements.

Reasoning, step by step

  1. Calculate P(A∣B)P(A \mid B) directly using set cardinalities: ∣A∩B∣/∣B∣|A \cap B| / |B|.
  2. Calculate the marginal probabilities P(A)P(A) and P(B)P(B) from the set sizes.
  3. Calculate the reverse conditional probability P(B∣A)P(B \mid A) using set cardinalities: ∣A∩B∣/∣A∣|A \cap B| / |A|.
  4. Substitute these values into Bayes' theorem formula.
  5. Verify that the result matches the direct calculation.

Example

The script states: 'In the nine equally likely outcomes, A contains five elements, B six, and their intersection two. Thus P(A∣B)=2/6=1/3P(A\mid B)=2/6=1/3, matching the Bayes calculation.'

Common misconceptions

  • Assuming that the probabilities must be calculated using frequencies rather than set counts in a uniform sample space.
  • Forgetting that the denominator in the direct calculation is the size of the conditioning event B, not the total sample space.

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