Given two events A and B. If odds against A are as 2 : 1 and those in favor of A∪B are as 3 : 1, then

A
$1 / 2 \leq P (B) \leq 3 / 4$
B
$5 / 12 \leq P (B) \leq 3 / 4$
C
$1 / 4 \leq P (B) \leq 3 / 5$
D
None of these

Solution:

$P (A) = \frac{1}{3}, P (A \cup B) = \frac{3}{4}$

Now,

$\begin{array}{l} P (A \cup B) = P (A) + P (B) - P (A \cap B) \leq P (A) + P (B) \\ \Rightarrow \frac{3}{4} \leq \frac{1}{3} + P (B) \Rightarrow \frac{5}{12} \leq P (B) \end{array}$

Again ,we have $B \subseteq A \cup B$

$∴ P (B) \leq P (A \cup B) = \frac{3}{4}$

Hence $\frac{5}{12} \leq P (B) \leq \frac{3}{4}$

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