Probabilities that neither sibling has inherited the trait, that Bob has inherited the trait 
Anna and Bob are siblings. There is probability $0.3$ that at least one of them inherited a certain genetic trait from their parents. With probability $0.2$, Anna has inherited the trait but Bob has not. (In other words, with probability $0.2$, Anna alone has inherited the trait.)
(a) Find the probability that neither one has inherited the trait.
(b) What is the probability that Bob has inherited the trait? Show your work or justify your answer.

Let "Anna has inherited"= A and "Bob has inherited"=B
$P(A \cup B)= 0.3$ or is it $P(A \cap B)=0.3$
$P(A)= 0.2$
have I got this part right?
for a) $P(\text{Neither})= 1-0.3 =0.7$
for b)
$P(B)= 1-(P(A)+P(\text{Neither}))$
$P(B)= 1-(0.2+0.7)$
$P(B)=0.1$
 A: You obtained the correct answers, but your justification for the first answer is not clear and your justification for the second answer is incorrect.
If we adopt your conventions that $A$ represents the event that Anna has inherited the trait and $B$ represents the event that Bob has inherited the trait, then $A \cup B$ is the event that at least one of them has inherited the trait, $A \cap B$ is the event that both of them have inherited the trait, and $A - B = A - (A \cap B)$ is the event that only Anna has inherited the trait.
We are given
\begin{align*}
\Pr(A \cup B) & = 0.3\\
\Pr(A - B) & = 0.2
\end{align*}
We are not given $\Pr(A)$.
The probability that neither sibling has inherited the trait is found by subtracting the probability that at least one of them has inherited the trait from $1$, which is 
$$\Pr((A \cup B)') = 1 - \Pr(A \cup B) = 1 - 0.3 = 0.7$$
The probability that Bob has inherited the trait is found by subtracting the probability that only Anna has inherited the trait from the probability that at least one of them has inherited the trait.  To see why, note that 
$$\Pr(A \cup B) = \Pr(A) + \Pr(B) - \Pr(A \cap B)$$
Since $\Pr(A - B) = \Pr(A) - \Pr(A \cap B)$, 
\begin{align*}
\Pr(B) & = \Pr(A \cup B) - \Pr(A) + \Pr(A \cap B)\\
       & = \Pr(A \cup B) - [\Pr(A) - \Pr(A \cap B)]\\ 
       & = \Pr(A \cup B) - \Pr(A - B)\\ 
       & = 0.3 - 0.2\\ 
       & = 0.1
\end{align*}
A: Yes, both your answers are correct. However, part of your working doesn't quite say what you intended.


*

*To answer your question at the start, it is $P(A\cup B)=0.3$; $A\cup B$ indicates "either $A$ or $B$ (or both)", i.e. at least one of them occurs. $P(A\cap B)$ would be the probability that both occur (you are not given enough information to deduce what this is).

*Also, for b) it should be $P(B)=1-P(A\cap B')-P(A'\cap B')$, that is $1-P(A\text{ alone})-P(\text{neither})$. Even though you wrote $P(A)$, you used the value of $P(A\text{ alone})$ so got the right answer.

