Given that $E$ and $F$ are events such that $P(E) = 0.8, P(F) = 0.7, P(E \cap F) = 0.6$. Find $P(\overline{E} | \overline{F})$.

$\frac{1}{3}$

The correct answer is Option (2) → $\frac{1}{3}$ ##

$P(\overline{E} | \overline{F}) = \frac{P(\overline{E} \cap \overline{F})}{P(\overline{F})} = \frac{P(\overline{E \cup F})}{P(\overline{F})} = \frac{1 - P(E \cup F)}{1 - P(F)} \quad \dots(i)$

Now, $ P(E \cup F) = P(E) + P(F) - P(E \cap F)$

$= 0.8 + 0.7 - 0.6 = 0.9$

Substituting value of $P(E \cup F)$ in (i):

$P(\overline{E} | \overline{F}) = \frac{1 - 0.9}{1 - 0.7} = \frac{0.1}{0.3} = \frac{1}{3}$