Probability Conditional Question Please Help I have a probability question which I would appreciate some help with.

a) Complete the values in the tree diagram
b) The probability that Terry will not be the champion is 0.58. Find the value of p
c) Given that Terry did not become the champion, find the probability she lost in the semi-final
My main struggle is with c as I understand it is a conditional probability question but personally I think the events are independent and thus P(A|B) = P(A) therefore 0.3 but the alternative is to say 15/29 by saying P(A and B) / P(B) = 0.3/0.58 = 15/29.
Any helpt would be greatly appreciated as to which working is correct. Thanks!!
 A: According to the definition of $\mathcal P(A \mid B),$
$$\mathcal P(A \cap B) = \mathcal P(A \mid B) \mathcal P(B).$$
Any intuition that leads you to a different conclusion is a mistaken intuition about conditional probabilities.
Dependence of one event on another in probability may arise because the other event has a causal effect on the first one, but that is not what defines dependence.
Dependence also is a two-way street: it happens when
$$  \mathcal P(A \cap B) \neq  \mathcal P(A)  \mathcal P(B),$$
and as you might guess from the symmetry of this formula, if $B$ is dependent on $A$ then $A$ is dependent on $B.$
If you want to think of this in more causal terms, consider
$\mathcal P(A \mid B) : (1 - \mathcal P(A \mid B))$
as the fair odds on a bet that $A$ will occur, with the condition that the bet is settled with these odds if $B$ occurs, but if $B$ does not occur the bet is canceled.
Suppose a friend says to you, "I bet you Terry will lose the semi-final. If she wins I pay you a dollar, if she loses you pay me two dollars." This is a favorable bet for you (and a bad bet for your friend) because the expected winnings of $1$ dollar with $0.7$ probability outweighs than the expected loss of $2$ dollars with $0.3$ probability.
But suppose your friend puts an extra condition on the bet--you have to wait until after the final before settling the bet, and if Terry wins the final the whole bet is canceled. You now have a much lower chance of winning, but just as big a chance of losing. In fact, is now a bad bet for you and a favorable bet for your friend. You should ask for better odds on the bet.
This corresponds to the fact that $\mathcal P(A \mid B) > \mathcal P(A)$.
A: B) The probability that Terry will not be the champion is
$$(1-p)+0.4p=1-0.6p$$
Solving,
\begin{align*}
    1-0.6p=0.58 \\
    p=0.7
\end{align*}
C) The answer is
$$\frac{1-p}{1-0.6p}=\frac{0.3}{0.58}=\frac{15}{29}$$
The events are not independent, since the event "Terry loses in the semi-final" is a subset of the event "Terry did not become the champion." Knowing the former, you also know that the latter occurs.
