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

Accepted Answer

Roots of $12x^2-7x+1=0$ are $x=\dfrac13,\dfrac14$. Take $P(A\mid B)=\dfrac13$, $P(B\mid A)=\dfrac14$.

From $P(A\mid B)=\dfrac{P(A\cap B)}{P(B)}=\dfrac13$: $P(B)=0.3$.

From $P(B\mid A)=\dfrac{P(A\cap B)}{P(A)}=\dfrac14$: $P(A)=0.4$.

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

$$P(\overline A\cup \overline B) = 1-P(A\cap B)=1-0.1=0.9,\qquad P(\overline A\cap \overline B)=1-P(A\cup B)=1-0.6=0.4$$

$$\frac{P(\overline A\cup \overline B)}{P(\overline A\cap \overline B)}=\frac{0.9}{0.4}=\frac94$$

Solutions · 1