In a soccer match probability that team $A$ scores first is $0.6$ and probability that team $B$ scores first is $0.35$, because in many matches neither teams scores the probability doesn't add up to $1$.
So the question is if team $B$ scores first, the prob. that team $A$ will win is $0.3$, given that $A$ won the match, find the prob. that $A$ did not score first.
So I've defined $P(A' \cap W)=0.3$ where $A'$ is not scoring first and $W$ is A won the match and the problem I've setup is $P(A'|W)$.
Unfortunately the correct $P(W)$ couldn't be found after I tried $P(B)*P(W)= 0.3$ making $P(W)$ the subject and using $P(B)=.35$
Answer = $0.179$ 3 s.f
Impertinent prior question if information is needed:
The prob. that A scores first and wins is $0.48$
The prob. that A scores first and does not win is: $P(A)*P(W)=P(A \cap W)$
$P(W)=0.8$
Therefore, $P(A)*(W')=P(A \cap W')=0.2*0.6=0.12$
$P(A \cap W')=0.12$
Best Answer
I wouldn't take the wording "if team $B$ scores first, the prob. that team $A$ will win is $0.3$" to mean that $\ P(A'\cap W)=0.3\ $, but that $\ P(W|B)=0.3\ $. On that understanding, we have $$ 0.3=P(W|B)=\frac{P(W\cap B)}{P(B)}= \frac{P(W\cap A')}{0.35} $$ because the each of the events $\ W\cap A'\ $ and $\ W\cap B\ $ occurs if and only if the other does. Therefore, \begin{align} P(W\cap A')&=0.105\ \ \text{ and}\\ P(A\cap W)&=0.48\ .\\ \end{align} Therefore $$ P(W)=P((W\cap A')\cup(A\cap W))=0.585 \ \ \text{, and}\\ P(A'|W)= \frac{P(W\cap A')}{P(W)}=\frac{0.105}{0.585}\approx0.179\\ $$