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
$\frac{1}{12}\le P\left(B\right)\le \frac{3}{4}$