Disjoint events

Question

Let $A$ and $B$ two disjoint events such that $P(A)=0.3$ and $P(B)=0.5$. Find the probability that

i)$A$ or $B$ ocurrs

ii)$A$ occur but not $B$

iii)repeat $i)$ and $ii)$ with $A$ and $B$ independent events

I was reviewing basics of probability, and this question left me with doubts. Since $A$ and $B$ are disjoints $P(A\cap B)=0$?

i)$$P(A\cup B)=P(A)+P(B)-P(A\cap B)=P(A)+P(B)-0=0.8$$

ii)Here I am confused by this "but", that means $P(A\cap B^c)$ or $$P(A|B^c)=\frac{P(A\cap B^c)}{P(B^c)}$$?

iii)$$P(A\cup B)=P(A)+P(B)-P(A\cap B)=P(A)+P(B)-P(A)*P(B)=0.65$$ I know that $$P(A|B)=^{independence}P(A)$$ but I don't remender if it's true for $$P(A|B^c)=P(A)$$?

Answer 1

Your first answer looks good: $\Pr(A \cup B) = 0.8$.

For your second answer, I believe they mean: what is the probability $A$ occurs and $B$ does not occur; or: $\Pr(A \cap B^c)$. Since $A \subset B^c$, it follows that $\Pr(A \cap B^c) = \Pr(A)=0.3$.

For independent events, we can multiply the probabilities:

\begin{align}&\Pr(A\cup B)= 1- \Pr(A^c \cap B^c)= 1- (0.7)(0.5) = 0.65.\\ &\Pr(A \cap B^c)= (0.3)(0.5)= 0.15.\end{align}

Answer 2

i) Yes, that is one way of stating two events are disjoint or mutually exclusive: $$P(A \cap B) = 0$$

ii) They mean the former. $P(A \cap B^c)$.

iii) $P(A|B^c) = P(A)$ when $B$ is independent of $A$. Here is a proof: $$P(A) = P(A | B)P(B) + P(A|B^c)P(B^c) = P(A)P(B) + P(A|B^c)(1 - P(B))$$ So $$P(A)(1 - P(B)) = P(A|B^c)(1 - P(B))$$