Calculating P(A^c \cup B^c) Given Conditional Probabilities | UPSC ISS 2024 Paper-1 | Problem-31

Описание: This explanation solves for \(12 \times P(A^c \cup B^c)\) given:
\(P(A) = \frac{1}{3}\), \(P(B) = \frac{1}{5}\), and
\(P(A|B) + P(B|A) = \frac{2}{3}\).

Using conditional probability formulas:
\[
P(A|B) = \frac{P(A \cap B)}{P(B)}, \quad P(B|A) = \frac{P(A \cap B)}{P(A)},
\]
we find:
\[
P(A \cap B) \times \left( \frac{1}{P(B)} + \frac{1}{P(A)} \right) = \frac{2}{3}.
\]
Substituting values provides \(P(A \cap B) = \frac{1}{2}\). Then,
\[
P(A^c \cup B^c) = 1 - P(A \cap B) = 1 - \frac{1}{2} = \frac{1}{2}.
\]
Multiplying by 12 yields:
\[
12 \times P(A^c \cup B^c) = 12 \times \frac{1}{2} = 6.
\]
This demonstrates how to manipulate event probabilities using conditional probabilities.

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