If A and B are two events such that , and P(A|B) and P(B|A) are the roots of the equation , then the value of is : [2025]

STUDY MATERIALS
COURSES
MORE
SUCCESS STORY
ADMISSION
STORE

Back

JEE ADVANCE

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

JEE MAIN

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

NEET

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

BITSAT

CLASS XI - XII

CLASS XI - XII

CBSE BOARD

CLASS IX

CLASS X

CLASS XI

CLASS XII

CLASS VI

CLASS VII

CLASS VIII

UP BOARD

CLASS IX
CLASS X
CLASS XI
CLASS XII

CLASS IX

CLASS X

CLASS XI

CLASS XII

CUET

CRASH COURSE

CRASH COURSE

Courses

CLASSROOM COURSES
ONLINE COURSES

ONLINE COURSES

More

Solution

Q.
If A and B are two events such that $P (A \cap B) = 0.1$ , and $P (A | B)$ and $P (B | A)$ are the roots of the equation $12 x^{2} - 7 x + 1 = 0$ , then the value of $\frac{P (\bar{A} \cup \bar{B})}{P (\bar{A} \cap \bar{B})}$ is : [2025]

1 $\frac{5}{3}$

2 $\frac{4}{3}$

3 $\frac{7}{4}$

4 $\frac{9}{4}$

Ans.
(4)

We have, $12 x^{2} - 7 x + 1 = 0$

$\Rightarrow x = \frac{1}{3}, \frac{1}{4}$

Let $P (\frac{A}{B}) = \frac{1}{3} a n d P (\frac{B}{A}) = \frac{1}{4}$

$\Rightarrow \frac{P (A \cap B)}{P (B)} = \frac{1}{3} and \frac{P (A \cap B)}{P (A)} = \frac{1}{4}$

$\Rightarrow P (B) = 0.3 and P (A) = 0.4$

$∴ P (A \cup B) = 0.3 + 0.4 - 0.1 = 0.6$

Now, $\frac{P (\bar{A} \cup \bar{B})}{P (\bar{A} \cap \bar{B})} = \frac{P (\bar{A \cap B})}{P (\bar{A \cup B})}$

$= \frac{1 - P (A \cap B)}{1 - P (A \cup B)} = \frac{1 - 0.1}{1 - 0.6} = \frac{9}{4}$ .

Want Us to Email you About Special Offers & Updates?