Aspire Faculty ID #11527 · Topic: NIMCET 2023 · Just now
NIMCET 2023

Given to events A and B such that odd in favour A are 2 : 1 and odd in favour of $A \cup B$ are 3 : 1. Consistent with this information the smallest and largest value for the probability of event B are given by

Solution

Given: Odds in favour of A = 2 : 1 
 So, $P(A) = \frac{2}{3}$ 
 Odds in favour of $A \cup B = 3 : 1$ 
 So, $P(A \cup B) = \frac{3}{4}$ 
 Let $P(B) = q$ and $P(A \cap B) = x$. 
 Using inclusion–exclusion: $P(A \cup B) = P(A) + P(B) - P(A \cap B)$ 
 So, $\frac{3}{4} = \frac{2}{3} + q - x$ $\Rightarrow x = \frac{2}{3} + q - \frac{3}{4}$ 

 Condition 1: $x \ge 0$ $\frac{2}{3} + q - \frac{3}{4} \ge 0$ $\Rightarrow q \ge \frac{1}{12}$ 

 Condition 2: $x \le P(A)$ $\frac{2}{3} + q - \frac{3}{4} \le \frac{2}{3}$ $\Rightarrow q \le \frac{3}{4}$ 

 Therefore possible range of $P(B)$ is: $\frac{1}{12} \le P(B) \le \frac{3}{4}$ 

 So minimum value = $\frac{1}{12}$ 
 Maximum value = $\frac{3}{4}$
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