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 $12x^2 – 7x + 1 = 0$, then the value of $\frac{P(\overline{A} \cup \overline{B})}{P(\overline{A} \cap \overline{B})}$ is:
- (1) $\frac{5}{3}$
- (2) $\frac{4}{3}$
- (3) $\frac{9}{4}$
- (4) $\frac{7}{4}$
Solution:
Roots of $12x^2 – 7x + 1 = 0$:
$12x^2 – 4x – 3x + 1 = 0 \Rightarrow 4x(3x-1) – 1(3x-1) = 0$
$x = \frac{1}{3}, \frac{1}{4}$.
Let $P(A|B) = \frac{1}{3}$ and $P(B|A) = \frac{1}{4}$ (order doesn’t affect final P(A) and P(B) calculation).
$P(A|B) = \frac{P(A \cap B)}{P(B)} \Rightarrow \frac{1}{3} = \frac{0.1}{P(B)} \Rightarrow P(B) = 0.3$.
$P(B|A) = \frac{P(A \cap B)}{P(A)} \Rightarrow \frac{1}{4} = \frac{0.1}{P(A)} \Rightarrow P(A) = 0.4$.
$P(A \cup B) = P(A) + P(B) – P(A \cap B) = 0.4 + 0.3 – 0.1 = 0.6$.
Numerator: $P(\overline{A} \cup \overline{B}) = P(\overline{A \cap B}) = 1 – P(A \cap B) = 1 – 0.1 = 0.9$.
Denominator: $P(\overline{A} \cap \overline{B}) = P(\overline{A \cup B}) = 1 – P(A \cup B) = 1 – 0.6 = 0.4$.
Ratio $= \frac{0.9}{0.4} = \frac{9}{4}$.
Ans. (3)