Question

Assertion (A) : Odd in favour of an events are $2 : 1$ & odd in favour of $A\cup B$ are 3 :1 then $\frac{1}{12} \le P\left(B\right) \le \frac{3}{4}$
Reason (R) : If $\left(A \cap B\right) \subset A$ then $P\left(A \cap B\right) \le P\left(A\right)$

• A. Both (A) & (R) are individually true & (R) is correct explanation of (A).
• B. Both (A) & (R) are individually true but (R) is not the correct (proper) explanation of (A).
• C. (A) is true but (R) is false.
• D. (A) is false but (R) is true.

Answer 1

Answer: (A)

Given, odd in favour of $A$ is $2 : 1$
$\therefore P\left(A\right) = \frac{2}{3}$
Odd in favour of $A \cup B$ is $3 : 1$
$\therefore P\left(A \cup B\right) = \frac{3}{4}$
Now $P\left(A \cup B\right) =P\left(A\right) +P\left(B\right) -P\left(A \cap B\right)$
$\Rightarrow \frac{3}{4} = \frac{2}{3} +P \left(B\right)-P\left(A \cap B\right)$
$\Rightarrow P\left(A \cap B\right) =P\left(B\right) +\frac{2}{3} -\frac{3}{4}$
$\Rightarrow P\left(A \cap B\right) =P\left(B\right) - \frac{1}{12}$.....(i)
$\Rightarrow P\left(B\right) \ge \frac{1}{12}$.....(*)
Again $P\left(A \cap B\right) \le P\left(A\right)$
$\Rightarrow P\left(B\right) - \frac{1}{12} \le \frac{2}{3}$(using (i))
$\Rightarrow P\left(B\right) \le \frac{2}{3} +\frac{1}{12} =\frac{3}{4}$.....(**)
By (*)&(**} we have
$\frac{1}{12} \le P\left(B\right) \le \frac{3}{4}$