First slide

Theorems of probability

Question

Let A, B, C be three mutually independent events. Consider the two statements S₁ and S₂.
S₁ : A and B $\cup$ C are independent.
S₂ : A and B $\cap$ C are independent.
Then

Moderate

A

both, S₁ and S₂ are true
B

only S₁ is true
C

only S₂ is true
D

neither S₁ nor S₂ is true

Solution

We are given that
$\begin{array}{l} P (A \cap B) = P (A) P (B) \\ P (B \cap C) = P (B) P (C) \\ P (C \cap A) = P (C) P (A) \\ P (A \cap B \cap C) = P (A) (B) P (C) \end{array}$
We have,
$\begin{matrix} P (A \cap (B \cap C)) = P (A \cap B \cap C) = P (A) P (B) P (C) \\ = P (A) P (B \cap C) \end{matrix}$
$\Rightarrow$ A and B a C are independent
Therefore, S₂ is true. Also,
$\begin{matrix} P [(A \cap (B \cup C)] = P [(A \cap B) \cup (A \cap C)] \\ = P (A \cap B) + P (A \cap C) - P [(A \cap B) \cap (A \cap C)] \\ = P (A \cap B) + P (A \cap C) - P (A \cap B \cap C) \\ = P (A) P (B) + P (A) P (C) - P (A) P (B) P (C) \\ = P (A) [P (B) + P (C) - P (B) P (C)] \\ = P (A) [P (B) + P (C) - P (B \cap C)] \\ = P (A) P (B \cup C) \end{matrix}$
Therefore, A and B $\cup$ C are independent

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