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Theorems of probability

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Question

$A and B are 2 events such that P (A) = \frac{3}{4} and P (B) = \frac{5}{8} . If a and b are the possible minimum and maximum values$
$of P (A \cap B), then the value of a + b is$

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Solution

$\begin{array}{l} A \cap B \subseteq B \\ P (A \cap B) \leq P (B) \\ P (A \cap B) \leq \frac{5}{8} - - - - (1) \\ Now P (A \cup B) = P (A) + P (B) - P (A \cap B) \\ P (A \cup B) \leq 1 \\ \Rightarrow P (A) + P (B) - P (A \cap B) \leq 1 \\ P (A \cap B) \geq \frac{3}{4} + \frac{5}{8} - 1 \\ i.e., P (A \cap B) \geq \frac{3}{8} - - - - - (2) \end{array}$
From (i) and (ii)
$\begin{array}{l} \frac{3}{8} \leq P (A \cap B) \leq \frac{5}{8} \\ ∴ x + y = \frac{3}{8} + \frac{5}{8} = 1 \end{array}$

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