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…

Question

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 $12 x^2-7 x+1=0$, then the value of $\frac{P(\bar{A} \cup \bar{B})}{P(\bar{A} \cap \bar{B})}$ is :

Accepted Answer

To solve this problem, start by considering the equation given for the probabilities:

$ 12x^2 - 7x + 1 = 0 $

The roots of this equation are:

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

Assume $ P(A \mid B) = \frac{1}{3} $ and $ P(B \mid A) = \frac{1}{4} $.

From the definitions of conditional probability, we have:

$ P(A \mid B) = \frac{P(A \cap B)}{P(B)} = \frac{1}{3} $

$ P(B \mid A) = \frac{P(A \cap B)}{P(A)} = \frac{1}{4} $

Given that $ P(A \cap B) = 0.1 $, we can use these equations to find $ P(B) $ and $ P(A) $:

From $ \frac{0.1}{P(B)} = \frac{1}{3} $, we find $ P(B) = 0.3 $.

From $ \frac{0.1}{P(A)} = \frac{1}{4} $, we find $ P(A) = 0.4 $.

Now, calculate $ P(A \cup B) $ using the formula for the union of two events:

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

The task is to find the value of:

$ \frac{P(\overline{A} \cup \overline{B})}{P(\overline{A} \cap \overline{B})} $

Using De Morgan's laws and the complements:

$ 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}) = 1 - P(A \cup B) = 1 - 0.6 = 0.4 $

Finally, compute the ratio:

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

Solutions · 1