Question
Let $A,B,C$ be three events such that $P(A \cap B) = 0,P(A \cup B) = 1,P(A \cap C) = \frac{1}{5}$ and $P(C) = \frac{7}{15}$, then $P(B \cap C)$ is equal to
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Solution
Sol. [Key Concept: $A,B,C$ with $P(A \cap \ B) = 0,\ P(A \cup \ B) = 1$: deduce the required probability.]
$$P(A \cup B \cup C) = 1,P(A \cap B \cap C) = 0$$
$$\begin{matrix} & & P(\text{ }A \cup \text{ }B) = P(\text{ }A) + P(\text{ }B) - P(\text{ }A \cap \text{ }B) \\ \Rightarrow & & P(\text{ }A) + P(\text{ }B) = 1 \\ & & P(\text{ }A \cup \text{ }B \cup C) = P(\text{ }A) + P(\text{ }B) + P(C) - (P(\text{ }A \cap \text{ }B) + P(\text{ }B \cap C) + P(C \cap \text{ }A)) + P(\text{ }A \cap \text{ }B \cap C) \\ & & \\ \Rightarrow & & 1 = 1 + \frac{7}{15} - \left( 0 + P(\text{ }B \cap C) + \frac{1}{5} \right) + 0 \\ \Rightarrow & & P(\text{ }B \cap C) = \frac{4}{15} \end{matrix}$$
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