Multiplicative property

Probability Level 2

Let A , B , C , D A, B, C, D be events in a sample space X X . It is known that:

  • P ( A ) > 0 P(A) > 0
  • P ( B ∣ A ) = 1 4 P(B \mid A) = \frac{1}{4}
  • P ( C ∣ B ∩ A ) = 1 3 P(C \mid B \cap A) = \frac{1}{3}
  • P ( D ∣ C ∩ B ∩ A ) = 1 2 P(D \mid C \cap B \cap A) = \frac{1}{2}

Then P ( A ) = m P ( D ∩ C ∩ B ∩ A ) P(A) = m P(D \cap C \cap B \cap A) with m ∈ Z m \in \mathbb{Z} . Submit m m .


The answer is 24.

Guillermo Templado by Guillermo Templado , 46, Spain

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1 solution

P ( D ∩ C ∩ B ∩ A ) P ( A ) = P ( D ∩ C ∩ B ∩ A ) P ( C ∩ B ∩ A ) ⋅ P ( C ∩ B ∩ A ) P ( B ∩ A ) ⋅ P ( B ∩ A ) P ( A ) = \frac{P(D \cap C \cap B \cap A)}{P(A)} = \frac{P(D \cap C \cap B \cap A)}{P(C \cap B \cap A)} \cdot \frac{P( C \cap B \cap A)}{P(B \cap A)} \cdot \frac{P( B \cap A)}{P(A)} = = P ( D ∣ C ∩ B ∩ A ) ⋅ P ( C ∣ B ∩ A ) ⋅ P ( B ∣ A ) = 1 2 ⋅ 1 3 ⋅ 1 4 = 1 24 ⟺ = P(D \mid C \cap B \cap A) \cdot P( C \mid B \cap A) \cdot P(B \mid A) = \frac{1}{2} \cdot \frac{1}{3} \cdot \frac{1}{4} = \frac{1}{24} \iff P ( A ) = 24 ⋅ P ( D ∩ C ∩ B ∩ A ) ⟺ m = 24 P(A) = 24 \cdot P(D \cap C \cap B \cap A) \iff m = 24

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