E,F independent; P(E F)=112, P( E F)=12.

Question

$E,F$ independent; $P(E\cap F)=\dfrac{1}{12}$, $P(\overline E\cap \overline F)=\dfrac12$. Find a value of $\dfrac{P(E)}{P(F)}$.

Accepted Answer

Let $P(E)=x,P(F)=y$. $xy=\dfrac1{12}$, $(1-x)(1-y)=\dfrac12 \Rightarrow x+y=\dfrac{7}{12}$.

Solving $12x^2-7x+1=0$ gives $x=\dfrac13$ or $\dfrac14$, with corresponding $y=\dfrac14$ or $\dfrac13$.

$$\frac{x}{y} = \frac{1/3}{1/4}=\frac43 \quad\text{(or } \frac34\text{)}$$

Solutions · 1