Q31: Independent E and F with $P(\bar E) = 0{\cdot}6$⁠, $P(E \cup F) = 0{\cdot}6$⁠: find $P(F)$ and $P(\bar E \cup \bar F)$

$\bar E$ is "E does not happen". An event and its complement have probabilities that add to 1:

This is the definition of independent events: the probability that both happen is the product of their probabilities. It turns the unknown $P(E \cap F)$ into $0{\cdot}4\,P(F)$⁠, so $P(F)$ is the only unknown left.

This is the addition rule. Put in $P(E \cup F) = 0{\cdot}6$⁠, $P(E) = 0{\cdot}4$ and $P(E \cap F) = 0{\cdot}4\,P(F)$⁠.

Then collect the $P(F)$ terms: $P(F) - 0{\cdot}4\,P(F) = 0{\cdot}6\,P(F)$⁠. Subtracting $0{\cdot}4$ from both sides leaves $0{\cdot}6 - 0{\cdot}4 = 0{\cdot}2$ on the left.

Turned round, this is $0{\cdot}6\,P(F) = 0{\cdot}2$⁠.

$P(F) = \frac{0{\cdot}2}{0{\cdot}6} = \frac{2}{6} = \frac13$⁠.

Divide both sides by $0{\cdot}6$⁠. Then multiply top and bottom by 10 to clear the decimals, and divide top and bottom by 2:

By De Morgan's law, "not E or not F" is the complement of "E and F":

Then use the complement rule and independence again. The product is $\frac{4}{10}\times\frac13 = \frac{4}{30} = \frac{2}{15}$⁠, and $1 - \frac{2}{15} = \frac{13}{15}$⁠.