More on Axiomatic Approach to Probability

Data Science and AI Lecture Series

By Bindeshwar Singh Kushwaha

Statement of the First Proof

Prove: \( P(A \cap B^c) = P(A) – P(A \cap B) \)

• This formula expresses the probability of \( A \) occurring without \( B \).
• It uses the complement rule and properties of set operations.

Proof of the First Statement

• The event \( A \) can be partitioned into two disjoint subsets:
  \[
  A = (A \cap B) \cup (A \cap B^c)
  \]
• Using the additivity property of probabilities:
  \[
  P(A) = P(A \cap B) + P(A \cap B^c)
  \]
• Rearranging gives:
  \[
  P(A \cap B^c) = P(A) – P(A \cap B)
  \]

Statement of the Second Proof

Prove: \( P(A^c \cap B) = P(B) – P(A \cap B) \)

• This formula expresses the probability of \( B \) occurring without \( A \).
• It also uses the complement rule and properties of set operations.

Proof of the Second Statement

• The event \( B \) can be partitioned into two disjoint subsets:
  \[
  B = (A \cap B) \cup (A^c \cap B)
  \]
• Using the additivity property of probabilities:
  \[
  P(B) = P(A \cap B) + P(A^c \cap B)
  \]
• Rearranging gives:
  \[
  P(A^c \cap B) = P(B) – P(A \cap B)
  \]

Example 4: Calculate \( P(A \cap B^c) \)

Given: \( P(A) = 0.6 \), \( P(A \cap B) = 0.2 \)

• By the formula:
  \[
  P(A \cap B^c) = P(A) – P(A \cap B)
  \]
• Substitute the values:
  \[
  P(A \cap B^c) = 0.6 – 0.2
  \]
• Simplify:
  \[
  P(A \cap B^c) = 0.4
  \]

Example 5: Calculate \( P(A^c \cap B) \)

Given: \( P(B) = 0.7 \), \( P(A \cap B) = 0.3 \)

• By the formula:
  \[
  P(A^c \cap B) = P(B) – P(A \cap B)
  \]
• Substitute the values:
  \[
  P(A^c \cap B) = 0.7 – 0.3
  \]
• Simplify:
  \[
  P(A^c \cap B) = 0.4
  \]

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