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

JEE Main · 2025 · 22 Jan 2025 · Shift 2 · Probability

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 ( A ˉ \cup B ˉ )}{P ( A ˉ \cap B ˉ )}

is:

Answer: (C)

Solving

12 x^{2} - 7 x + 1 = 0

gives

x = \frac{1}{3}

x = \frac{1}{4}

, the two conditional probabilities.

Take

P (A ∣ B) = \frac{1}{3} \Rightarrow P (B) = \frac{0.1}{1/3} = 0.3

, and

P (B ∣ A) = \frac{1}{4} \Rightarrow P (A) = \frac{0.1}{1/4} = 0.4

P (A \cup B) = P (A) + P (B) - P (A \cap B) = 0.4 + 0.3 - 0.1 = 0.6.

By De Morgan's laws,

P (\overset{ˉ}{A} \cup \overset{ˉ}{B}) = 1 - P (A \cap B) = 0.9, P (\overset{ˉ}{A} \cap \overset{ˉ}{B}) = 1 - P (A \cup B) = 0.4.

\frac{P ( A ˉ \cup B ˉ )}{P ( A ˉ \cap B ˉ )} = \frac{0.9}{0.4} = \frac{9}{4} .

(Assigning the roots the other way round gives the same values of

P (A) + P (B)

, hence the same ratio.)

JEE Main · 2025 · 22 Jan 2025 · Shift 2 · Single correct option