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]

Question

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]

Smart Achievers · Accepted Answer

(4)

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

$P (A | B)$

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

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

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 $\frac{9}{4}$

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Solution

Q. If A and B are two events such that P(A∩B)=0.1, and P(A|B) and P(B|A) are the roots of the equation 12x2–7x+1=0, then the value of P(A¯∪B¯)P(A¯∩B¯) is : [2025]

1 53

2 43

3 74

4 94