If A and B are two events such that P(A cap B | vedclass

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(C) Given the quadratic equation $12x^2 - 7x + 1 = 0$. Solving for $x$: $12x^2 - 4x - 3x + 1 = 0$ $\Rightarrow 4x(3x - 1) - 1(3x - 1) = 0$ $\Rightarro

Vedclass · Accepted Answer

$\frac{9}{4}$

Vedclass · Answer

(C) Given the quadratic equation $12x^2 - 7x + 1 = 0$. Solving for $x$: $12x^2 - 4x - 3x + 1 = 0$ $\Rightarrow 4x(3x - 1) - 1(3x - 1) = 0$ $\Rightarrow (4x - 1)(3x - 1) = 0$. So,the roots are $x = \frac{1}{3}$ and $x = \frac{1}{4}$. Let $P(A \mid B) = \frac{1}{3}$ and $P(B \mid A) = \frac{1}{4}$. Using the definition of conditional probability: $P(A \mid B) = \frac{P(A \cap B)}{P(B)} = \frac{0.1}{P(B)} = \frac{1}{3} \Rightarrow P(B) = 0.3$. $P(B \mid A) = \frac{P(A \cap B)}{P(A)} = \frac{0.1}{P(A)} = \frac{1}{4} \Rightarrow P(A) = 0.4$. Now,$P(A \cup B) = P(A) + P(B) - P(A \cap B) = 0.4 + 0.3 - 0.1 = 0.6$. Using De Morgan's Laws: $P(\overline{A} \cup \overline{B}) = P(\overline{A \cap B}) = 1 - P(A \cap B) = 1 - 0.1 = 0.9$. $P(\overline{A} \cap \overline{B}) = P(\overline{A \cup B}) = 1 - P(A \cup B) = 1 - 0.6 = 0.4$. Therefore,$\frac{P(\overline{A} \cup \overline{B})}{P(\overline{A} \cap \overline{B})} = \frac{0.9}{0.4} = \frac{9}{4}$.

If $A$ and $B$ are two events such that $P(A \cap B) = 0.1$,and $P(A \mid B)$ and $P(B \mid A)$ 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:

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