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

Accepted Answer

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

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

The roots of this equation are:

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

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

From the definitions of conditional probability, we have:

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

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

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:

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

Using De Morgan's laws and the complements:

$P(\vec{A}\cup \vec{B})=P(\vec{A\cap B})=1-P(A\cap B)=1-0.1=0.9$

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

Finally, compute the ratio:

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

Solutions · 1